IN  MEMORIAM 
FLORIAN  CAJOR1 


e  cu<^ 


ESSENTIALS    OF 

CALCULUS 


BY 


E.  J.  TOWNSEND,  Ph.D.  (Gottingen) 

Professor  of  Mathematics 
University  of  Illinois 


and 

G.  A.  GOODENOUGH,  M.E.  (Illinois) 

Associate  Professor  of  Mechanical  Engineering 
University  of  Illinois 


NEW  YORK 

HENRY  HOLT  AND  COMPANY 

1910 


Copyright,  1910 

BY 

HENRY   HOLT  AND  COMPANY 


NortoootJ  $reBg 

.7.  S.  Cashing  Co.  — Berwick  &  Smith  Co. 

Norwood,  Mass.,  U.S.A. 


T(o 


6*i 


PREFACE 

In  the  preparation  of  this  volume,  the  authors  have  had  in 
mind  the  needs  of  those  colleges  and  technical  schools  in  which 
the  time  devoted  to  calculus  is  limited  to  a  three-hour  course  for 
a  year,  or  perhaps  to  a  five-hour  course  for  two  terms. 

The  usual  division  of  the  subject  into  differential  and  integral 
calculus  has  been  largely  disregarded.  By  the  arrangement 
adopted,  the  student  is  early  led  by  easy  steps  into  simple  prac- 
tical applications  of  the  calculus ;  and  the  more  difficult  topics 
are  postponed  until  late  in  the  course. 

The  theory  of  limits  has  been  used  exclusively  in  the  develop- 
ment of  fundamental  principles.  Throughout  the  book  much 
emphasis  is  placed  upon  the  applications  of  the  calculus  to  prac- 
tical problems.  Only  such  knowledge  of  physics  on  the  part 
of  the  student  is  assumed  as  is  usually  included  in  an  elementary 
course  in  that  subject.  Some  problems  are  introduced  that  show 
the  use  of  calculus  in  discussing  well-known  applications  to 
physical  and  engineering  phenomena.  Such  problems  are  so 
stated,  however,  as  to  require  no  technical  knowledge  on  the  part 
of  the  student.  The  applications  to  geometry  are  such  as  are 
essential  and  usually  to  be  found  in  a  first  course  in  calculus. 

In  the  selection  of  material,  the  authors  have  departed  some- 
what from  the  traditional  course.  Many  topics  usually  included 
in  calculus  have  been  entirely  omitted  or  greatly  reduced  in 
extent.  Thus,  but  little  attention  has  been  given  to  special 
methods  of  integration ;  and  reduction  formulas  for  integration, 
order  of  contact,  envelopes,  etc.,  have  been  omitted  entirely. 
On  the  other  hand,  some  parts  of  the  text  have  been  extended 
beyond  the  usual  limits.  Functions  of  two  or  more  variables, 
because  of  their  importance  in  physics,  have  been  discussed  more 
fully  than  usual.     Special  stress  has  been  laid  on  the  summation 


iv  PREFACE 

process,  and  by  numerous  examples  from  physics  the  student 
is  drilled  in  the  choice  of  proper  elements  and  in  the  setting  up 
of  definite  integrals.  Attention  is  called  to  the  treatment  of 
exact  and  inexact  differentials ;  a  subject  of  first  importance 
in  the  physical  applications  of  calculus,  but  one  that  usually 
receives  little  or  no  consideration. 

The  authors  take  this  occasion  to  express  their  obligations 
to  their  colleagues  at  the  University  of  Illinois  and  elsewhere 
for  their  helpful  suggestions  in  the  preparation  of  this  book,  and 
to  the  publishers  for  their  cooperation  in  making  its  typography 
of  high  grade.  The  authors  are  under  special  obligations  to 
Professor  H.  L.  Eietz  and  to  Dr.  A.  R.  Crathorne  for  their  assist- 
ance in  seeing  the  book  through  the  press. 

E.    J.    TOWN  SEND. 

G.   A.    GOOBENOUGH. 
Univkrsity  of  Illinois, 
July,  1910. 


CONTENTS 

CHAPTER   I 
CONSTANTS,  VARIABLES,  FUNCTIONS 

ARTICLE  PAGE 

1.  Constants,  Variables 1 

2.  Functions 2 

3.  Fundamental  Problems  of  Calculus 3 

4.  Slope  of  a  Tangent 3 

5.  Speed  of  a  Falling  Body 6 

6.  Given  a  Derivative  to  find  the  Function 7 

7.  Area  under  a  Curve 7 

8.  Functional  Notation 10 

9.  Graphs  of  Functions 11 

10.  Definition  of  a  Limit .         .         .12 

11.  Infinity 14 

12.  Continuity  of  Functions 15 

13.  Laws  of  Operation  with  Limits 18 

14.  Limit  of  a  Monotone  Function 20 

15.  Indeterminate  Forms 21 

CHAPTER   II 
DERIVATIVES  OF  ALGEBRAIC  FUNCTIONS 

16.  Increments 27 

17.  Definition  of  a  Derivative 28 

18.  Conditions  for  a  Derivative 30 

19.  Geometrical  and  Physical  Significance  of  the  Derivative          .         .  31 
20     General  Theorems  on  Differentiation .32 

21.  Derivative  of  a  Constant 32 

22.  Derivative  of  a  Variable  with  Respect  to  Itself         ....  32 

23.  Derivative  of  a  Sum 33 

24.  Derivative  of  a  Function  plus  a  Constant 33 

25.  Derivative  of  a  Product 34 

26.  Derivative  of  a  Constant  times  a  Function 35 

27.  Derivative  of  a  Quotient 36 

28.  Derivative  of  a  Constant  by  a  Variable 37 

v 


vi  CONTENTS 

ARTICLK  PAGE 

29.  Derivative  of  un 38 

30.  Algebraic  Functions 39 

31.  Derivative  of  a  Function  of  a  Function .41 

32.  Derivatives  of  Inverse  Functions 42 

33.  Derivative  of  One  Function  with  Respect  to  Another,  when  Both 

are  Functions  of  a  Common  Variable 44 


CHAPTER   III 
ELEMENTARY  APPLICATIONS  OF  DERIVATIVES 

34.  Slope  of  a  Curve .         .         .47 

35.  Derived  Curves 49 

36.  Rolle's  Theorem 52 

37.  Law  of  the  Mean 53 

38.  Tangents  and  Normals  to  Curves 55 

39.  Length  of  Tangent,  Normal ;  Subtangent,  Subnormal     ...  57 

40.  Tan  ft  cot  ^ 58 

41.  Length  of  Tangent,  Normal ;   Polar  Subtangent,  and  Polar  Sub- 

normal   60 

42.  Speed  and  Acceleration     .  61 

43.  Angular  Speed  and  Acceleration 62 

44.  Miscellaneous  Applications  of  Derivatives       .        .        .        .        .64 

CHAPTER   IV 
THE  DIFFERENTIAL  NOTATION 

45.  The  Derivative  as  a  Rate 69 

46.  Differentials      ..." 70 

47.  Kinematic  Interpretation  of  Differentials 72 

48.  Differentiation  with  Differentials 75 

49.  Differentiation  of  Implicit  Functions 77 

50.  Applications  of  Rates  and  Differentials 79 

CHAPTER  V 
DIFFERENTIATION  OF  TRANSCENDENTAL  FUNCTIONS 

51.  Transcendental  Functions .         .83 

52.  Differentiation  of  sin  u 83 

53.  Differentiation  of  cos  u 84 

54.  Differentiation  of  tan  u 84 

55.  Differentiation  of  cot  u 85 


CONTENTS 


Vll 


ARTICLE 

56.  Differentiation  of  sec  u  and  esc  u 

57.  Inverse  Trigonometric  Functions 

58.  Differentiation  of  arc  sin  u  and  arc  cosw 

59.  Differentiation  of  arc  tan  u  and  arc  cot  u 

60.  Differentiation  of  arc  sec  u  and  arc  esc  u 

61.  Exponential  and  Logarithmic  Functions 

62.  Differentiation  of  au  and  eu 

63.  Differentiation  of  loga  u    . 

64.  Differentiation  of  log  u 

65.  Logarithmic  Differentiation 


85 
87 
88 
89 
90 
92 
94 
95 
95 
96 


CHAPTER   VI 


INTEGRATION 


66.  Anti-derivatives  and  Integrals 

67.  General  Theorems    . 

68.  The  Integral   (un  du 

69.  Fundamental  Integrals 

70.  Integration  by  Inspection. 

71.  Integration  by  Substitution 

72.  Character  of  the  Integration  Process 


101 
102 

103 

104 
105 
108 
110 


CHAPTER   VII 
SIMPLE  APPLICATIONS  OF  INTEGRATION 

73.  Curves  having  Given  Properties 113 

74.  Rectilinear  Motion 114 

75.  Rotation  about  a  Fixed  Axis 116 

76.  Motion  of  a  Projectile 116 

77.  Harmonic  Motion     .         . 118 

78.  Motion  in  a  Resisting  Medium  .         .         ...         .         .        .119 

79.  Physical  Problems  involving  Exponential  Functions        .        .        .  122 


CHAPTER   VIII 
SUCCESSIVE  DIFFERENTIATION  AND  INTEGRATION 

80.  Definition  of  the  nth  Derivative 126 

81.  Successive  Differentiation  of  Implicit  Functions       ....  127 

82.  Geometrical  and  Physical  Interpretations  of  the  Second  Derivative  128 

83.  Successive  Integration 129 


viii  CONTENTS 


84.  Maxima  and  Minima 132 

85.  Applications  of  Maxima  and  Minima 136 

CHAPTER   IX 

CURVES 

86.  Concavity 142 

87.  Points  of  Inflexion 144 

88.  Asymptotes,  Rectangular  Coordinates 145 

89.  Singular  Points 149 

90.  Curve  Tracing 152 

91.  Definition  and  Measure  of  Curvature 157 

92.  Radius  of  Curvature.     Center  of  Curvature 158 

93.  Radius  of  Curvature,  Parametric  Representation  ....  160 

94.  Roulettes  and  Involutes 102 

95.  Curvature  of  Involutes 163 

96.  Evolute  of  a  Curve 165 

CHAPTER   X 
DEFINITE  INTEGRALS 

97.  Definition  of  a  Definite  Integral 169 

98.  Elementary  Properties  of  Definite  Integrals 171 

99.  Change  of  Limits 172 

100.    The  Definite  Integral  as  the  Limit  of  a  Sum 173 

401.   Importance  of  the  Summation  Process 176 

102.  Geometrical  Representation  of  a  Definite  Integral          .        .        .  178 

103.  Definite  Integrals  of  Discontinuous  Functions.     Infinite  Limits  of 

Integration 179 

CHAPTER  XI 
APPLICATIONS   OF  INTEGRATION  TO  GEOMETRY  AND  MECHANICS 

104.  Plane  Areas,  Rectangular  Coordinates 184 

105.  Plane  Areas,  Polar  Coordinates .  186 

106.  Volumes  of  Solids  of  Revolution 188 

107.  Volumes  determined  by  the  Summation  of  Slices  .         .         .         .  190 

108.  Lengths  of  Curves,  Rectangular  Coordinates 192 

109.  Lengths  of  Curves,  Polar  Coordinates 195 

110.  Areas  of  Surfaces  of  Revolution 196 

111.  Mean  Value 198 


CONTENTS  ix 

ARTICLE  pAGE 

112.  Work  of  a  Variable  Force 200 

113.  Work  of  Expanding  Gases 203 

CHAPTER   XII 
SPECIAL  METHODS   OF  INTEGRATION 

114.  Integration  by  Parts 207 

115.  Integration  of  Rational  Fractions 209 

116.  Integration  of  Functions  containing  Radicals          ....  213 

117.  Integration  of  Special  Trigonometric  Functions      ....  218 

118.  Use  of  a  Table  of  Integrals 221 

119.  Approximate  Determination  of  Integrals 223 

120.  Simpson's  Rules 224 

121.  Mechanical  Integration 227 

CHAPTER   XIII 
FUNCTIONS  OF  TWO   OR  MORE   VARIABLES 

122.  Definition  of  a  Function  of  Several  Variables         .         .         .         .  231 

123.  Partial  Derivatives 232 

124.  Interchange  of  Order  of  Differentiation 236 

125.  Total  Derivatives 237 

126.  Total  Differentials 242 

127.  Differentiation  of  Implicit  Functions 244 

128.  Exact  and  Inexact  Differentials 245 

CHAPTER   XIV 
MULTIPLE  INTEGRALS,  APPLICATIONS 

129.  Multiple  Integrals 254 

130.  Plane  Areas  by  Double  Integration  ;  Rectangular  Coordinates      .  257 

131.  Plane  Areas  by  Double  Integration  ;  Polar  Coordinates         .         .  259 

132.  Volumes  by  Triple  Integration  ;  Rectangular  Coordinates     .         .  261 

133.  Volumes  by  Triple  Integration  ;  Polar  Coordinates        .         .         .  265 

134.  Additional  Examples  involving  Multiple  Integrals          .         .         .  267 

135.  Mass.     Mean  Density 271 

136.  First  Moments.     Centroids 273 

137.  General  Theorems  relating  to  Centroids 276 

138.  Second  Moment.     Radius  of  Gyration 280 

139.  General  Theorems  relating  to  Second  Moments      ....  282 

140.  Illustrative  Examples 286 


CONTENTS 


CHAPTER  XV 
INFINITE   SERIES 

ARTICLE  PAGE 

141.  Fundamental  Definitions 291 

142.  Tests  of  Convergence 293 

143.  Power  Series 295 

144.  Maclaurin's  Expansion  of  a  Function  in  a  Power  Series         .         .  296 

145.  Taylor's  Expansion 298 

146.  Taylor's  Theorem.     Maclaurin's  Theorem 301 

147.  Integration  and  Differentiation  of  Series 303 

148.  Use  of  Series  in  Computation .  306 

149.  Approximation  Formulas 308 

150.  Maxima  and  Minima  of  Functions  of  a  Single  Variable          .         .  311 

151.  Evaluation  of  Indeterminate  Forms 312 

152.  Analytic  Condition  for  a  Singular  Point 317 


COURSE   IN   CALCULUS 


GREEK   ALPHABET 


Letters 

Letters 

Capitals 

Lower 

Case 

Names 

Capitals 

Lower 
Case 

Names 

A 

a 

Alpha 

N 

V 

Nu 

B 

ft 

Beta 

E 

$ 

Xi 

r 

y 

Gamma 

0 

o 

0  micron 

A 

8 

Delta 

n 

7T 

Pi 

E 

e 

Epsilon 

p 

P 

Rho 

Z 

i 

Zeta 

2 

<T 

Sigma 

H 

V 

Eta 

T 

T 

Tau 

© 

6 

Theta 

Y 

V 

Upsilon 

I 

i 

Iota 

3> 

4> 

Phi 

K 

K 

Kappa 

X 

X 

Chi 

A 

X 

Lambda 

* 

* 

Psi 

M 

A* 

Mu 

o 

(O 

Omega 

Xll 


CALCULUS 


CHAPTER  I 
FUNDAMENTAL   NOTIONS   AND    DEFINITIONS 

1.  Constants,  variables.  In  the  applications  of  mathematics  to 
physical  problems,  we  meet  with  such  magnitudes  as  velocity, 
force,  mass,  length,  area,  volume,  etc.  The  measure  of  a  magni- 
tude is  expressed  by  means  of  a  number,  represented  by  a  figure 
or  letter,  which  denotes  the  ratio  that  the  given  magnitude  bears 
to  some  standard  magnitude  of  the  same  kind  adopted  as  a  unit. 
For  the  purposes  of  calculation,  it  is  the  number  that  is  of  funda- 
mental importance.  For  the  sake  of  brevity,  however,  we  shall 
often  speak  of  "a  velocity  v"  or  "an  area  A"  etc.,  instead  of 
using  the  longer  but  perhaps  more  precise  expressions  "  a  velocity 
whose  measure  is  v  units  "or  "  an  area  whose  ratio  to  the  standard 
unit  of  area  is  A,"  etc. 

In  any  particular  discussion,  the  magnitudes  and  consequently 
the  corresponding  numbers  may  or  may  not  change.  A  number 
that  remains  unchanged  is  called  a  constant.  A  symbol  is  then 
said  to  represent  a  constant  when  it  denotes  but  one  value  in  a 
given  discussion.  A  symbol  that  satisfies  this  condition  is  itself 
often  called  a  constant. 

When  a  number  is  permitted  to  assume  different  values  in  the 
same  discussion,  it  is  called  a  variable.  A  symbol  is  said  to  repre- 
sent a  variable  if  it  has  assigned  to  it  different  values  in  the  same 
discussion.  Here  again  it  is  usual  to  speak  of  the  symbol  itself 
as  the  variable. 

For  example,  suppose  a  body  falls  from  rest.  The  law  that 
gives  the  relation  between  the  time  t  and  the  distance  s  through 
which  the  body  falls  is  expressed  by  the  equation 

s=lgt*.  (1) 

1 


2  FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 

Here  the  number  g  is  a  constant  denoting  the  acceleration  at  a 
point  on  the  earth's  surface.  On  the  other  hand,  the  time  t  and 
the  distance  s  are  variables,  whose  mutual  relation  is  determined 
by  the  given  equation. 

Again,  in  the  equation  of  the  circle 

(x-2y+(y-3y  =  r>,  (2) 

2  and  3  are  constants  that  determine  the  center  of  the  circle,  and 
r  is  a  constant  denoting  the  length  of  the  radius.  However,  x 
and  y  are  variables  whose  corresponding  values,  as  determined 
by  the  given  equation,  fix  the  various  positions  of  the  generating 
point.  To  the  constant  r  we  may  assign  any  value  at  pleasure, 
but  when  it  has  been  once  assigned  it  must  remain  unchanged, 
so  long  as  the  circle  remains  fixed.  A  constant  of  this  character 
is  called  an  arbitrary  constant  or  parameter. 

2.  Functions.  In  each  of  the  illustrations  given  in  the  preced- 
ing article,  it  will  be  observed  that  the  variables  involved  stand 
in  an  intimate  relation  to  each  other.  For  example,  in  the  law 
of  falling  bodies,  s  and  t  are  so  related  that  to  any  value  assigned 
to  t  there  corresponds  a  definite  value  of  s.  Moreover,  to  every 
positive  value  that  we  may  give  s,  the  given  relation  between  s 
and  t  determines  corresponding  values  of  t.  Again,  in  the  equation 
of  the  circle  the  variables  x  and  y  are  related  in  a  similar  manner. 
Other  illustrations  of  such  relations  between  variables  are  familiar 
to  the  student  from  his  study  of  elementary  mathematics  and 
physics.  Whenever  such  a  relation  exists  between  two  variables, 
we  say  that  the  variables  are  connected  by  a  functional  relation, 
or  that  the  one  is  a  function  of  the  other.  We  may  then  define  a 
function  as  follows : 

If  two  variables  are  so  related  that  for  each  value  that  may  be 
assigned  to  the  one  there  are  determined  one  or  more  definite  values 
of  the  other,  the  second  variable  is  said  to  be  a  function  of  the  first. 

The  variable  to  which  we  may  assign  arbitrarily  chosen  values 
is  called  the  independent  variable.  It  is  also  frequently  referred 
to  simply  as  the  variable  or  argument.  The  variable  which  is  thus 
determined,  that  is,  the  function,  is  sometimes  called  the  dependent 
variable. 


Arts.  2-4]  SLOPE   OF   A   TANGENT  3 

A  function  may  depend  for  its  value  upon  two  or  more  inde- 
pendent variables.  For  example,  the  area  of  an  ellipse  is  a 
function  of  the  lengths  of  the  major  and  the  minor  axes ;  the 
volume  of  a  gas  is  a  function  of  the  temperature  and  of  the 
pressure  to  which  the  gas  is  subjected.  Such  functions  are  said  to 
be  functions  of  two  variables.  For  the  present  we  shall  consider 
only  functions  of  a  single  variable ;  later,  some  of  the  properties 
of  functions  of  two  or  more  variables  will  be  discussed. 

3.  Fundamental  problems  of  the  calculus.  In  his  study  of  ele- 
mentary mathematics  and  in  his  everyday  experiences,  the  student 
has  frequently  had  occasion  to  deal  with  magnitudes  that  change. 
In  many  instances  the  changes  are  abrupt  and  sometimes  periodic. 
For  example,  the  market  price  of  any  commercial  product  changes 
abruptly  from  time  to  time.  When  money  is  placed  at  compound 
interest,  the  amount  of  interest  is  usually  added  to  the  principal 
at  certain  intervals.  Among  physical  phenomena,  on  the  other 
hand,  we  encounter  numerous  illustrations  of  changes  which  are 
evidently  continuous.  Thus  the  pressure  of  the  atmosphere  varies 
with  the  altitude,  but  the  change  is  gradual,  not  abrupt ;  the  speed 
of  a  railway  train  starting  from  a  station  changes  continuously 
until  the  maximum  speed  is  reached;  the  pressure  of  a  liquid 
upon  a  vertical  wall  increases  continuously  with  the  depth.  Many 
other  illustrations  of  both  continuous  and  discontinuous  changes 
will  occur  to  the  student. 

Problems  involving  discontinuous  changes  are  dealt  with  for 
the  most  part  by  the  ordinary  processes  of  arithmetic  and  algebra. 
Problems  that  involve  continuous  changes  require  more  powerful 
mathematical  methods  and  these  are  the  special  province  of  the 
calculus.  To  illustrate  the  methods  of  the  calculus  and  in  a 
general  way  give  the  student  some  notion  of  the  scope  of  the  sub- 
ject and  of  the  class  of  problems  to  be  considered,  a  few  typical 
problems  of  fundamental  importance  are  here  introduced. 

4.  Problem  i.  Slope  of  a  tangent.  Required  the  angle  that 
the  tangent  to  a  given  curve  at  a  point  P1  (Fig.  1)  makes  with 
the  X-axis.  To  make  the  problem  concrete  let  the  equation 
of  the  curve  be  y  —  3  V#,  and  let  the  abscissa  of  PY  be  x  =  4. 

Through  the  given  point  Px  let  a  secant  line  be  drawn  cutting 


4  FUNDAMENTAL  NOTIONS   AND   DEFINITIONS     [Chap.  I. 

the  curve  in  a  second  point  P.     Denoting  by  Ax  the  increase  PXQ, 
in  the  abscissa  between  Pl  and  P,  by  Ay  the  corresponding  increase 

Y 


Fig.  1. 

QP  of  the  ordinate,  and  by  6  the  angle  that  the  secant  makes 
with  the  X-axis,  we  have 

P,Q     Ax 

As  the  point  P  is  made  to  approach  the  fixed  point  Ply  the  secant 
approaches  as  a  limiting  position  the  tangent  at  Pl  and  the  angle 
6  approaches  the  angle  <f>.    Hence,  by  taking  Ax  smaller  and  smaller 

and  calculating  the  ratio  — U  for  the  various  values  of  Ax,  we  can 

Ax 

approximate  more  and  more  closely  the  value  of  tan  <f>.  The 
following  table  gives  the  calculated  values  for  different  assumed 
values  of  Ax. 


Ax 

±y 

Ay 

AX   ■ 

2. 

1.34847 

0.67424 

1. 

0.708204 

0.70820 

0.1 

0.074537 

0.74537 

0.01 

0.007495 

0.74953 

Using  this  arithmetical  method,  we  can  arrive  at  an  approximate 
value  of  tan  </>.     The  method,  however,  is  tedious  and  the  result  is 


Art.  4]  SLOPE   OF  A  TANGENT  5 

at  best  an  approximation.     What  we  need  is  the  limiting  value 

of  —  as  Ax  approaches  the  value  zero.     This  limit  can  be  found 

Aa; 

by  considering  the  equation  of  the  curve. 

Let  (xu  yi)  be  the  coordinates  of  the  given  point  Pv  The  co- 
ordinates of  the  point  P  are  then  (xx  +  Aa?,  yx  +  Ay).  Hence  from 
the  equation  of  the  curve,  y  =  3 VaJ,  we  have 

y1  =  S^/xl7  (1) 


yx  +  Ay  =  3  V^  +  Aa;.  (2) 

By  subtraction,  we  have 

Ay  =  3(Vx1-\-  Ax  —  Va?i),  (3) 

and  therefore  it  follows  that 

Ay  _  g  Va?!  +  Ax  —  Va^  ,^ 

Aa;  Aa; 

Rationalizing  the  numerator,  we  obtain 

Ay 3  Aa; 3 ,^ 

Aa;     Ax(Vxx  +  Aa;  +  Vxi)      V  a?x  +  Aa;  +  Vol 

As  Aa;  approaches  zero,  the  expression  — == =  approaches 

■\/xl  -f  Aa;  -+-  Va;x 

_,  and  since  at  the  same  time  tan  0,  which  is  given  by  the  ratio 

2Va;x 

— y. ,  approaches  tan  <f>,  we  infer  that  tan  <f>  = •     We  have  then 

Aa;'    FF  ^  r     2Va>! 

3    ' 
the  result  that  for  a%  =  4,  tan  <f>  =  =  0.75. 

2V4 

The  method  here  developed  has  the  added  advantage,  that  the 

result  is  general  and  the  slope  of  the  tangent  at  any  point  can  be 

found  when  once  we  know  the  abscissa  of  the  point.     All  we  need 

to  do  is  to  substitute  the  given  value  of  x  in  the  general  formula 

q 

— —  •     The  difficulty  of  applying  this  method  to  all  problems  is 

2Vx 

that  in  most  cases  that  arise  it  is  not  easy  to  find  the  limiting 

value  of  _^.     In  the  subsequent  chapters  of  the  calculus  we  shall 

Aa; 


6  FUNDAMENTAL   NOTIONS   AND  DEFINITIONS     [Chap.  I. 

develop  methods  by  means  of  which  this  limit,  known  as  the 
derivative  of  y  with  respect  to  a?,  may  be  obtained  directly  from 
the  given  functional  relation  between  x  and  y. 

5.  Problem  2.  Speed  of  a  falling  body.  To  determine  the  average 
speed  of  a  moving  body  during  a  given  time,  we  simply  divide  the 
distance  traveled  by  the  time  occupied.  Thus,  an  eighteen-hour 
train  from  Chicago  to  New  York  has  an  average  speed  of  905  -f- 18 
=  50T5 g-  miles  per  hour.  The  speed  at  any  particular  instant  is  not 
so  easily  determined.  The  obvious  way  to  find  this  speed  approxi- 
mately is  to  take  small  intervals  of  time,  say  5  seconds,  1  second, 
-j-1^  second,  etc.,  immediately  following  the  instant  in  question,  and 
divide  the  distance  traversed  in  the  assumed  time  interval  by  that 
time  interval.  The  result  is  the  mean  speed  for  the  time  interval, 
and  it  is  evident  that  the  smaller  the  chosen  interval  is  taken  the 
more  nearly  the  quotient  gives  the  instantaneous  speed  at  the  be- 
ginning of  the  interval.  Such  a  method  is,  however,  open  to  the 
criticism  mentioned  in  the  previous  problem  ;  namely,  no  matter 
how  small  the  interval  taken,  the  result  is  merely  an  approxima- 
tion. To  obviate  this  objection  we  make  use  of  the  definite  law 
which  the  motion  follows  and  as  in  problem  1  find  the  value  of 
the  limit  involved. 

As  a  concrete  case,  let  us  consider  the  motion  of  a  body  falling 
in  a  vacuum.  From  physics,  it  is  known  that  s  =  \  gt2,  where  s 
denotes  the  distance  traversed  in  t  seconds  starting  from  rest,  and 
g  is  a  constant  whose  value  is  approximately  32.2.  Suppose  we 
wish  to  find  the  speed  at  the  end  of  tx  seconds.  The  distance  s1 
traversed  in  the  time  tx  is  given  by 

If  M  denotes  the  assumed  time  interval  immediately  following  t}, 
and  As  denotes  the  distance  traversed  in  this  interval  of  time,  we 
have 

*1  +  A8  =  $gr(«l  +  AQ8.  (2) 

From  (1)  and  (2)  we  obtain 

=  gtAt  +  \g(MY. 


Arts.  5-7]  AREA   UNDER   A   CURVE  7 

As 

Therefore,  we  have  —  =  gt1+±gAt.  (3) 

Equation  (3)  gives  the  mean  speed  for  the  time  At,  and  it  is  seen 
that  this  speed  is  a  variable  magnitude  depending  upon  the  as- 
sumed time  interval  At. 

As  At  is  taken  smaller,  this  mean  speed  for  the  interval  ap- 
proaches the  instantaneous  speed  at  the  end  of  tx  seconds.  How- 
ever, as  At  becomes  smaller,  the  expression  gtx-\-±g  At  approaches 
the  fixed  value  gtv  We  conclude,  therefore,  that  gtx  is  the  actual 
speed  at  the  end  of  tx  seconds.  At  the  end  of  3  seconds,  for  ex- 
ample, the  speed  is  32.2  x  3  =  96.6  feet  per  second,  and  at  the  end 
of  10  seconds  322  feet  per  second.  When  we  have  learned  how  to 
find  the  derivative  of  the  function  s  with  respect  to  t,  that  is,  the 

A9 

limit  of  the  ratio  — ,  directly  from  the  functional  relation  between 

At 

s  and  t,  the  process  here  indicated  will  be  very  much  simplified. 

6.  Problem  3.     Given  a  derivative,  to  find  the  original  function. 

In  problems  1  and  2  we  saw  the  importance  of  being  able  to  find 
from  the  functional  relation  between  two  variables  the  derivative 
of  the  one  with  respect  to  the  other,  that  is,  the  limiting  values  of 

—  and  —  •  In  the  one  case,  the  result  gave  the  slope  of  the  tan- 
Ax  At  &  * 

gent  to  a  curve,  and  in  the  other,  the  speed  of  a  moving  body  at  a 

particular  instant.     It  is  often  equally  as  important  to  be  able  to 

3 

solve  the  inverse  problem ;  that  is,  if  we  have  given as  the 

2V# 
slope  of  the  tangent  to  a  curve,  or  gt  as  the  speed  of  a  moving 
body,  it  is  of  value  in  certain  discussions  to  be  able  to  say  that  the 
curve  in  question  is  given  by  the  functional  relation  y  =  3Vx  or 
in  the  other  case  that  the  law  of  motion  for  the  body  is  expressed 
by  the  equation  s  =  \  gt2.  A  process  of  the  calculus  known  as  in- 
tegration enables  us  to  solve  such  problems. 

7.  Problem  4.  Area  under  a  curve.  One  of  the  most  important 
problems  in  calculus,  in  fact,  one  of  the  problems  that  led  to  the 
invention  of  the  calculus,  is  that  of  finding  the  area  between  a 
given  curve,  the  X-axis  and  two  given  ordinates.     To  take  a  con- 


8 


FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 


crete  example,  let  us  attempt 
to  find  the  area  between  the 
parabola  x  =  y2,  the  X-axis,  the 
origin,  and  the  ordinate  x  =  3. 
See  Fig.  2. 

Let  the  part  of  the  X-axis 
between  x  =  0  and  x  =  3  be 
divided  into  n  equal  parts,  each 
denoted  by  Ax.  At  the  subdivi- 
sions An  A.2,  etc.,  let  ordinates 
be  erected  and  rectangles  be 
constructed  as  shown  in  the 
figure.  The  abcissas  of  the 
points  Ax,  A2,  As,  etc.,  are  Ax, 
2  Ax,  3  Ax,  etc.,  and  the  altitudes 
of  the  successive  rectangles  are 

0,  (Ax)2,  (2  Ax)2,  etc.     Hence  the  sum  of  the  areas  of  the  rectangles 

is 

4.  =  0  •  Aa>  +  Ax  (Ax)2  +  Ax  (2  Ax)2  + [ ...  +  Aaj  (n  -  1  Ax)2 
=  (Ax)3[l2  +  22  +  32  +  ...  +(n  _l)*]. 

From  algebra,*  we  have 


*■  X 


Fig.  2. 


i»+2«+3«+  -  +(»-i)'=n(n-iyn-i) 

Hence,  we  have  from  (1) 

r       V.       ;  6 


(1) 

(2) 


n  •  Ax 


2  j>  .  Ax)2  -  3  (n  •  Ax)  Ax  +  (Ax)2  \ 
6  J 


(3) 


But,  it  will  be  remembered  that 

n  •  Ax  =  segment  (L4, 
Equation  (3)  then  becomes 


3. 


Ar  =  3  A8-9Ax+(Ax)2Y 


(4) 


*  Rietz  &  Crathorne's  College  Algebra,  p.  37,  Ex.  4. 


Art.  7]  AREA   UNDER   A   CURVE  9 

Equation  (4)  gives  the  sum  of  the  areas  of  the  rectangles  for  any 
assumed  value  of  Ax.  As  Ax  is  taken  smaller  and  smaller  and 
the  number  of  rectangles  is  correspondingly  increased,  the  area  Ar 
approaches  as  a  limit  the  area  A  under  the  curve,  that  is,  the  area 
OP^.      However,  from  (4)  it  follows  that  as  Ax  approaches 

zero,  Ar  has  the  limiting  value  — - —  =  9,  and  hence  we  have  as 
a  result  ^4  =  9. 

This  problem  illustrates  a  large  and  important  class  of  problems 
considered  in  calculus,  namely,  those  requiring  for  their  solution 
the  summation  of  an  indefinitely  large  number  of  indefinitely 
small  elements.  All  problems  in  areas  and  volumes,  and  many 
of  the  applications  to  mechanics  and  mathematical  physics,  are  of 
this  character. 

The  preceding  problems  have  been  selected  as  typical  of  the 
kind  of  problems  to  be  discussed  later  and  as  indicating  some- 
what the  class  of  problems  to  which  the  methods  of  the  calculus 
particularly  apply.  The  student  will  have  observed  that  each 
of  these  problems  depends  for  its  solution  upon  the  finding  the 
limiting  value  of  some  function.  However,  the  methods  employed 
in  these  examples  for  finding  the  limit  are  not  of  great  value, 
because  they  become  too  complicated  and  tedious  when  applied 
to  any  but  the  simplest  cases.  For  example,  in  the  last  problem, 
we  were  able  to  find  the  limit  easily  because  the  sum  of  the 
squares  of  the  first  n  integers  happens  to  be  known.  If  the  curve 
were  given  by  the  equation  y  =  V#,  we  should  have 

^r=(A^)3(Vl+V2+V3-h  •••  +Vw^T), 

and  the  solution  would  be  much  more  difficult.  In  the  succeeding 
chapters  we  shall  develop  methods  for  accomplishing  this  same 
purpose  much  more  directly  and  easily.  In  the  meantime  it  is 
essential  that  we  call  attention  to  some  of  the  fundamental  prop- 
erties of  functions  and  the  laws  of  limits  as  applied  to  the  prob- 
lems that  we  shall  need  to  consider. 

EXERCISES 

1.  By  the  method  of  problem  4  find  the  area  between  the  X-axis,  the 
line  y  =  2  x,  and  the  ordinate  x  =  5. 

2.  Find  the  slope  of  the  tangent  to  the  curve  y  =  x2  at  the  point  x  =  3. 


10  FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 

12 

3.  Find  the  slope  of  the  tangent  to  the  curve  y  —  —  at  the  point  x  =  6. 

x 

4.  If  a,  body  is  projected  vertically  upward  with  an  initial  speed  v0,  the 
relation  between  the  distance  s  and  the  time  t  is  s  =  v0t  —  §  gt2.  Find  an 
expression  for  the  speed  at  the  end  of  t\  seconds.  If  v0  =  150,  find  the  speed 
at  the  end  of  4  seconds. 

8.  Functional  notation.  As  we  have  seen  (Arts.  2,  4,  5),  the 
relation  between  a  variable  and  a  function  of  it  is  expressed  by 
an  equation;  as  y  =  3^/x,  s  =  ^gt2.  However,  even  when  the 
analytic  relation  between  the  independent  and  dependent  varia- 
bles is  known,  it  is  convenient  to  have  a  general  symbol  that  shall 
stand  for  the  function  as  expressed  in  terms  of  its  variable.  For 
this  purpose  the  variable  is  inclosed  in  a  parenthesis  and  some 
letter  is  prefixed.  Thus,  a  function  of  x  may  be  denoted  by  such 
symbols  as  the  following: 

f{x),  F{x),  <f>(x),  $(x)f  etc., 

which  are  read  "/  function  of  x  "  (or  simply  "/of  x  "),  "^function 
of  x"  etc.  It  is  to  be  understood  that  the  letters  /  F,  <£,  \J/,  etc., 
are  symbols  of  functionality  and  not  factors.  Whenever  two  or 
more  different  functions  are  employed  in  the  same  discussion,  a 
different  symbol  must  be  used  for  each. 

If  a  function  depends  upon  two  variables  instead  of  one,  it  may 
be  expressed  symbolically  by  f(x,  y),  <f>(x,  y),  etc. 

When,  in  connection  with  any  discussion,  we  have  once  defined 
/(#),  then  /(a)  denotes  the  same  expression  with  x  replaced  by  a. 
In  a  similar  way  we  may  form  functions  f(t),  /(4),  f(x  +  h)f  etc. 
Thus,  if  we  have 

f(x)  =  3x2  +  7»  +  9, 
it  follows  that 

/(*)  =  3e»+7*  +  9, 

/(4)  =  3.42  +  7.  4  +  9  =  85, 
fix  +  h)  =  3(x  +  hf  +  lix  +  h)  +  9. 

Instead  of  replacing  the  variable  by  another  variable  or  by  a 
constant,  we  may  replace  it  by  any  function  of  the  same  variable 
or  of  a  different  variable.     Thus  we  may  write 

/[*(*)],  F[f(y)l  <A(sin  0),  etc. 


Arts.  8,  9]  GRAPHS   OF  FUNCTIONS  11 

Given,  for  example,  the  function 

F(x)  =  x2-3x  +  7, 

we  have 

.F(sin  x)  =  sin2  x  —  3  sin  x  +  7. 

EXERCISES 

1.  Given  /(x)  =  x3-  6  x2  +  5x-  14.     Find  the  values  of  /(4),  /(0), 
/(  —  3).     Write  the  expression  for  /(sin  0)  and  for/(x  +  h). 

2.  If  0(x)  =  5 x2  —  2 x  +  8,  write  expressions  for  0(x2),  0(—  x3),  0(tan  0), 
0(0),  0(f). 

3.  If  F(d)  =  cos  0,  find  the  value  of  F(0),  W-Y  W-Y  ^00- 

4.  If  f(z)  -  x5  -  4  x3  +  x,  show  that  /(  -  x)  =  -/(x) . 

5.  If  /(x)  =  x4  -  6  x2  +  1 ,  show  that  /(  -  x)  =/  (x) . 

6.  Given  F(x)  =  3  -  Vxand/(y)  =  y2  +  4.    Find2?*[/(y)]  and/[JF(x)], 
also  /[/(y)J  and  F[F(x)l 

7.  Given  y  =/(x)  =    "*"g)  and  0  =fiy)  =/[/(«)]  ;  express  2  as  a  func- 

J.  —  x 
tion  of  x. 

8.  If  0(x)  =  2  x2  —  1,  show  that  0(cos  0)  =  cos  2  0. 


9.    If  0(x)  =  2  xv  1  -  x2,  show  that  0(sin  0)  =  0(cos  0)  =  sin  2  0. 

10.  If  /(x)  =logax,   show  that  f{x)  -f(y)  =/(*).   and  /(x)  +/(y) 

11.  If/(0)  =  cos0,  showthat/(0)=/(-0)  =  -/(7r  +  0)  =  -/(7r-0). 

12.  If/(x)  =sinx  and  0(x)  =  cosx,  find/(x).  <p(y)±  0(x)   /(y). 

13.  If /(x)  =  6«  *  show  that/(x)  •  /(?/)  =/(x  +  y). 


14.  If/(x)  =  Vl-x2,  find /(sin  0)  and /(cos  0). 

15.  If/(x)  =  Vl  +  x2,  find  /  (tan  0). 

16.  If  f{x)  =  tan  x,  find    &&T&L1   , 

!+/(*)/(*/) 

9.   Graphs  of  functions.     In  accordance  with  the  principles  of 
analytic  geometry,  a  geometric  interpretation  may  be  given   to 


*  The  symbol  e  is  used  to  denote  the  base  of  the  natural  system  of  loga- 
rithms, namely,  2.718  •••.  Log  x  indicates  the  logarithm  of  x  to  the  base  e. 
When  any  other  base  is  used,  that  fact  will  be  indicated  by  a  subscript, 
as  logax,  logi030. 


12  FUNDAMENTAL    NOTIONS   AND  DEFINITIONS     [Chap.  I. 

the  relation  of  a  function  to  its  variable.  If  the  values  of  the 
independent  variable  be  laid  off  as  abscissas,  the  corresponding 
values  of  the  function  may  be  used  as  ordinates.  The  assemblage 
of  the  extremities  of  these  ordinates  is  called  the  graph  of  the 
function.  It  is  to  be  noted  that  the  ordinates  of  the  points  on 
the  graph  and;  not  the  graph  itself  represent  the  values  of  the 
function.  It  is  assumed  that  the  student  is  already  familiar  with 
the  graphical  representation  of  functions  from  his  study  of  ana- 
lytic geometry. 

10.  Definition  of  a  limit.  The  general  notion  of  a  limit  is 
perhaps  fairly  clear  from  the  illustrative  problems,  Arts.  4-7. 
We  must,  however,  formulate  a  definition  with  sufficient  accuracy 
that  it  may  be  made  the  basis  of  a  mathematical  investigation. 
For  this  purpose,  let  us  consider  two  variables,  one  a  function  of 
the  other ;  say,  y  =f(x). 

Let  the  independent  variable  x  vary  in  such  a  manner  that  it 
differs  less  and  less  numerically  from  some  constant  a,  that  is,  in 
such  a  way  that  the  numerical  value  of  x  —  a,  written  \x  —  a\, 
becomes  and  remains  less  than  any  positive  constant  however 
small.  We  say  then  that  x  approaches  a,  and  indicate  that  fact 
by  writing  x  =  a,  which  is  to  be  read  "  x  approaches  a." 

As  the  independent  variable  x  approaches  the  constant  a  the 
dependent  variable  or  function  f(x)  assumes  a  corresponding  set 
of  values.  When  x  becomes  very  nearly  equal  to  a,  these  values 
oif(x)  may  at  the  same  time  become  very  nearly  equal  to  some 
constant,  say  A.  Moreover,  it  may  occur  that  we  may  not  only 
make  f(x)  differ  as  little  as  we  please  from  A  by  taking  a  value 
of  x  sufficiently  near  to  a,  but  that  it  will  remain  at  least  as 
close  to  A  for  all  values  of  x  that  lie  between  a  and  this  chosen 
value  of  x.  When  these  conditions  are  fulfilled,  we  call  A  the 
limit  of  the  function  f(x)  as  x  approaches  a. 

The  problem  of  falling  bodies,  Art.  5,  furnishes  a  good  illus- 
tration. The  function  gtx  +  ^gAt  not  only  may  be  made  to 
differ  as  little  as  we  please  from  gtx  by  taking  At  sufficiently  small, 
but  as  At  assumes  any  value  still  closer  to  zero  the  function  differs 
less  from  gtv  Hence  we  were  justified  in  speaking  of  gtx  as  the 
limit  of  gti  +  ^  g  At,  as  At  =  0. 


Art.  10]  DEFINITION   OF   A  LIMIT  13 

We  may  now  define  a  limit  as  follows : 

If  a  constant  A  can  be  found  such  that,  as  x  approaches  a,  f(x)  —  A 
becomes  and  remains  less  numerically  than  any  constant  however 
small,  then  A  is  called  the  limit  of  f(x)  as  x  approaches  a. 

If  the  function  f(x)  has  the  limit  A,  we  indicate  the  fact  by 
writing 

l  m=A.  a) 

x±a 

Since  by  the  definition  of  a  limit /(a?)  can  be  made  to  differ  from 
A  as  little  as  we  please  by  the  proper  selection  of  x,  we  may  also 
write 

f(X)  =  A  +  *(x),  (2) 

where  e(x)  can  be  made  as  small  as  we  choose  by  taking  x  suffi- 
ciently near  to  a.  Equations  (1)  and  (2)  are  equivalent  and  are 
merely  different  forms  of  expression  for  the  same  relation.  In 
some  demonstrations  form  (2)  will  be  found  the  more  convenient. 

Ex.     Consider  the  limit  of  the  function 

4 


/(*) 


x2  +  l 

as  x  =  2.     By  taking  the  value  of  x  sufficiently  near  2,  we  may  make  the 
value  of  f{x)  differ  as  little  as  we  please  from  i.     Hence,  we  have 

L        4  4 


£C  =  i 


a2  +  1      5 


In  the  preceding  discussion  of  limits  nothing  has  been  said 
about  the  value  of  f(x)  for  x  =  a.  The  limit  depends  for  its 
value  upon  the  values  of  f(x)  for  x  very  close  to  x  =  a,  or,  as  we 
frequently  say,  "  in  the  neighborhood  of  a,"  and  it  is  not  affected 
by  the  value  that/(#)  takes  for  x  =  a.  As  we  shall  see  later  the 
value  that  the  function  takes  for  x  =  a,  that  is,  f(a),  may  or  may 
not  be  equal  to  the  limit  of  f(x)  as  x  approaches  a. 

To  determine  whether  a  function  has  a  limit  as  the  variable 
approaches  a  definite  value,  we  must  consider  all  values  of  the 
function  in  the  neighborhood  of  the  limiting  value  of  the  variable. 
In  other  words,  a  limit  of  the  values  of  the  function  must  exist 
as  the  variable  approaches  its  limit  from  either  direction,  and 


14 


FUNDAMENTAL   NOTIONS  AND   DEFINITIONS     [Chap.  I. 


these  two  limiting  values  of  the  function  must  be  equal, 
following  example  serves  as  an  illustration. 


The 


Fig.  3. 


Ex.    Given  the  function 

y  =  arc  tan  - ,    the   graph 

x 
of  which  is  shown  in  Fig. 
3.    When  the  independent 
variable    is    restricted    to 
positive  values,  we  obtain 

the  limit  -  as  x  =  0  ;  when 

2 
negative   values    only  are 

considered,  we  get  —  -  as 

x  ±-  0.  Hence  the  function 
cannot  be  said  to  have  a 
limit  as  x  =  0. 


11.  Infinity.  In  the  discussion  of  limits  thus  far,  we  have 
considered  only  the  case  in  which  the  independent  variable  ap- 
proaches a  definite  number.  The  student  is  familiar  from  his 
study  of  algebra  and  geometry  with  the  case  in  which  the  inde- 
pendent variable  increases  without  limit.  For  example,  the  area 
of  a  circle  is  the  limit  of  the  area  of  a  regular  inscribed  polygon 
as  the  number  of  sides  increases  without  limit.  Again,  the  value 
of  an  infinite  series  is  obtained  by  taking  the  limit  of  the  sum  of 
a  finite  number  of  terms  as  the  number  of  terms  increases  with- 
out limit.  We  say  in  such  cases  that  the  independent  variable 
becomes  infinite,  and  express  that  fact  by  writing  x  =  oo,  n  =  oo, 
etc.  Here  the  symbol  =  is  not  to  be  understood  in  the  ordinary 
sense  of  "is  equal  to/'  but  rather  in  the  sense  of  "becomes,"  and 
the  above  expressions  should  be  read  "x  becomes  infinite,"  "n  be- 
comes infinite,"  etc.  Instead  of  the  independent  variable,  the 
function  may  become  infinite.  This  may  occur  when  the  varia- 
ble also  becomes  infinite,  or  when  it  approaches  a  definite  number. 
Thus  ar1  becomes  infinite  as  x  =  0.     While  it  is  customary  to 

write     L   ar1  =  oo  ,  the  function  cannot  be  said  to  have  a  limit, 
x  =  0 

because  it  does  not  approach  any  definite  number  however  large 

that  number  may  be. 

When  a  function  has  the  limit  zero,  it  is  often  spoken  of  as  an 

infinitesimal. 


Arts.  11,  12]  INFINITY  15 

The  introduction  of  the  concept  infinity  extends  the  use  of 
limits  so  as  to  include  a  large  number  of  important  applications, 
of  which  the  following  are  illustrations. 

Ex.  1.  An  air  pump  is  used  to  remove  air  from  an  inclosed  space.  At 
each  stroke  of  the  piston,  part  of  the  air  is  removed,  and  in  consequence  the 
density  of  the  air  remaining  is  diminished.  Let  d  denote  the  original 
density,  d„  the  density  after  n  strokes  of  the  piston,  v  the  volume  of  the 
inclosed  space,  and  v'  the  volume  of  the  pump  cylinder.  From  physics,  we 
have 

Here  the  density  dn  is  a  function  of  the  number  of  strokes.     If  n  is  increased 
without  limit,  we  have,  since  the  denominator  is  greater  than  unity, 

L  d         -0. 


K)* 


Hence,  as  the  variable  increases  without  limit,  the  function  approaches  0  as  a 
limit.  With  a  perfect  pump  we  may  therefore  make  the  density  as  small  as 
we  please  by  taking  the  number  of  strokes  sufficiently  large. 

Ex.  2.    The  work  done  by  a  gas  in  expanding  from  a  volume   V\  to  a 
volume  v  according  to  the  law  pvk  =  const,  is  given  by  the  expression 


Pivi'f    1 LI 

fc-lUi*"1     v*-1! 


If  the  expansion  proceeds  without  limit,  that  is,  if  v  =  oo  ,  the  expression  for 
the  work  becomes 

r  Pig*  [A Ll  -  pm*  [~— ol  =  PlVl 

^xfc-iU*-1   t?*-1  J   fc-iUi*-1     J   *-i' 

In  this  case  the  function  approaches  a  definite  limit  as  the  variable  increases 
without  limit. 

Ex.  3.    If    the    gas    expands    according    to    Boyle's    law,    pv  =  const., 
the  expression  for  the  work  done  is  piVi  log  — .     If,  in  this  case,  the  volume 
v  increases  without  limit,  we  have  for  the  work  done 
L   piVi  (log  v  —  log  tfi)'=  oo  ; 

that  is,  the  function  representing  the  work  done  also  increases  without  limit. 

12.   Continuity  of  functions.     We  have  seen  that  the  value  of 
a  limit  depends  upon  the  values  of  the  function  in  the  vicinity  of 


16  FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 

the  limiting  value  of  the  variable,  and  not  upon  the  value  of  the 
function  at  that  point.  In  some  cases  the  limit  is  the  same  as  the 
value  of  the  function  at  the  limiting  point,  and  in  some  cases  it 
is  not.  If  the  two  values  are  equal,  the  function  is  said  to  be 
continuous  at  the  point  in  question.  If,  on  the  other  hand,  the 
limit  is  different  from  the  value  of  the  function  at  the  limiting 
point,  or  if  for  any  reason  the  function  has  no  limit,  the  function 
is  said  to  be  discontinuous  at  that  point.  The  condition  that  f(x) 
shall  be  continuous  for  x  =  a  is  then  given  by  the  equation 

L  f(x)  =/(«). 

x  =  a 

As  we  shall  see,  continuity  is  a  very  important  property  of  the 
functions  to  be  discussed  later.  In  fact,  we  shall  confine  our- 
selves for  the  most  part  to  functions  having  this  property.  The 
following  examples  will  serve  to  make  clear  the  distinction  be- 
tween continuous  and  discontinuous  functions. 

Ex.  1.    The  function 

is  continuous  for  all  values  of  t ;  for,  we  have 

L  lgt*  =  lgt0i=f(t0), 
t=t0 

where  t0  is  any  definite  value  of  t. 
Ex.  2.     Consider  the  function 

for  values  in  the  neighborhood  of  the  origin.     We  have 

L   1 


which  is  another  way  of  saying  that  as  x  =  0,  the  value  of  the  function  in- 
creases without  limit,  and  hence  has  no  limit.  The  function  is  therefore  dis- 
continuous for  x  =  0.  A  function  always  has  a  discontinuity  for  any  value 
of  the  variable  for  which  the  function  becomes  infinite. 

We  have  a  similar  condition  in  Boyle's  law,  which  is  expressed  by  the 

equation  pv  =  k  or  p  =  k  -  •     If  the  volume  v  be  diminished  indefinitely, 

v 

the  pressure  increases  without  limit  and  hence  has  a  discontinuity  for 
v=0. 


Art.  12] 


CONTINUITY   OF   FUNCTIONS 


17 


Ex.  3.  Suppose  a  given  mass  of  ice  at  a  temperature  below  32°  F.  to  be 
heated.  As  heat  is  applied,  the  temperature  of  the  ice  rises,  and  the  quantity 
of  heat  Q  is  a  function  of  the  temperature  t.  So  long  as  t  remains  below  the 
melting  point  the  function  is  continuous.  But  when  the  melting  point  32°  F. 
is  reached,  a  quantity  of  heat  represented  by  AB,  Fig.  4,  is  absorbed  without 


7) 


-prater. 


Steam 


Boiling 


any  change  in  temperature.  Like- 
wise, when  the  water  reaches  the 
boiling  point,  heat  represented  by 
CD  is  absorbed  without  change  of 
temperature.  At  these  particular 
temperatures  the  function  is  there- 
fore discontinuous. 

In  the  preceding  discussion, 
we  have  spoken  only  of  the 
continuity  of  a  function  at  a 
single  point.  If  a  function  is 
continuous  at  all  points  of  an 
interval,  then  it  is  said  to  be  continuous  throughout  the  interval. 

In  plotting  a  graph,  it  is  of  great  assistance  to  know  that  the 
graph  represents  a  continuous  function  and  has  a  definite  direction 
at  each  point;  for  then  we  need  only  to  locate  points  of  the  graph 
sufficiently  dense  to  give  the  general  outline  and  to  draw  through 
these  points  a  continuous  curve.  Since  it  is  obviously  impossible 
to  locate  all  of  the  points  of  a  curve,  this  is,  in  fact,  the  only  way 
in  which  a  graph  can  be  drawn.  In  case  the  function  has  a  dis- 
continuity, care  must  be  taken  to  locate  points  sufficiently  dense 
in  the  neighborhood  of  the  discontinuity  to  determine  its  character. 


Fig.  4. 


EXERCISES 

1.  Which  of  the  trigonometric  functions  have  points  of  discontinuity  ? 

2.  Find  the  values  of  x  for  which  the  function  — '  x  +  * —  is  discontinuous. 

a;2-5x  +  6 

3.  If  f(x)  and  <f>(x)  are  integral  rational  functions,  under  what  conditions 
is  the  function  IS^l  discontinuous  ? 

4.  Examine  for  continuity  the  functions 

1 


(a)  y  =  arc  tan 


(b)y  = 


1  +e* 


18  FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 

13.  Laws  of  operation  with  limits.  In  this  section,  we  shall 
state  without  proof  *  several  properties  of  limits  that  will  be 
utilized  in  later  discussions.  These  are  given  in  the  following 
theorems. 

Theorem  I.  The  limit  of  a  constant  times  a  function  is  equal  to 
the  constant  times  the  limit  of  the  function  ;  that  is, 

L  c>f(x)  =  c-Lf(x). 
x=a  x=a 

Theorem  II.  If  each  of  two  functions  approaches  a  limit,  the 
limit  of  the  sum  (or  difference)  of  the  functions  is  equal  to  the  sum 
(or  difference)  of  the  limits;  that  is, 

L  [/(a>)  +  <KaO]  =  £   /(*)  +  -£  *(»)• 
x=a  x=a  x  =  a 

Theorem  III.  If  each  of  tivo  functioyis  has  a  limit,  the  limit  of 
the  product  of  the  functions  is  equal  to  the  product  of  the  limits; 
that  is, 

L   lf(x)  •  *(*)]  =  L  f(x)  .  L  <f>(x). 
x=a  x=a  x  =  a 

When  Theorem  III  is  extended  to  the  product  of  a  finite  num- 
ber n  of  equal  functions,  we  have  the  following : 

Corollary.  If  a  function  has  a  limit,  the  limit  of  the  nth  power 
of  the  function  is  equal  to  the  nth  power  of  its  limit. 

Theorem  IV.  If  each  of  two  functions  has  a  limit,  the  limit  of 
the  quotient  of  the  functions  is  equal  to  the  quotient  of  the  limits,  pro- 
vided the  limit  of  the  denominator  is  different  from  zero;  that  is, 

f(x)       L   AX) 

x  =  a4>(x)         L   <t>(x)    x  =  a 
x  =  a 

Theorem  V.     If  F(y)  is  a  continuous  function  of  y,  and  y  is 

any  function  of  x,  say  y  =  <f)(x),  such  that  L  <fi(x)=b, 

x  =  a 

then  L   F[^(*)]  =  F[  L  <£(>')]• 

x±a  x=a 


*  For  formal  proof  of  these  theorems  see  First  Course  in  Calculus,  or 
Bietz  and  Crathorne's  College  Algebra. 


Art.  13] 


LAWS   OF   OPERATION  WITH  LIMITS 


19 


The  following  example  will  illustrate  the  use  of  this  theorem : 
Ex.   Find  the  limit  L  log  (s2+  2  x  +  1). 

a- =2 

Since  the  logarithm  is  a  continuous  function  for  all  values  different  from 
zero,  we  may  apply  the  above  theorem  and  write 

L  log  (z2  +  2  x  +  1)  =  log  L  (x2  +  2  x  +  1)  =  log  9. 
a?=2  »=2 

In  fact,  we  may  always  write 

L  log0(x)=log  L  <p(x), 

x  =  a  x  =  a 

provided  the  limit  X  0(x)  is  different  from  zero. 


Theorem  VI.  If  a  given  function  lies  between  the  values  of  two 
other  functions  which  approach  a  common  limit,  the  given  function 
approaches  the  same  limit. 

Thus,  if  <j>(x)  Sl/(x)  ^  if,(x), 


and 

then 


L  <f>(x)  =  L  ij/(x)  =  A; 

x  =  a  x  =  a 

Lf(x)  =  A. 
x  =  a 


Ex.   Show  that  L  ?^JL  =  i,  and  L  ^^-  =  1. 

0=0     6  0=0     0 

In  Fig.  5,  an   arc  AD  is   described   with   a  radius   OA  =  r,  and   angle 
A  OD  =  0.     Then  we  have,  if  0  is  expressed  in  radians, 

arc  AD  =  r0, 

AB  =  r  sin  0, 

AG  =  rtsm  0. 
From  geometry  * 

AB<AD<AC, 
or  r  sin  0  <  r0  <  r  tan  0. 

Dividing  by  r  sin  0,  we  get 

1<  -^-<sec0. 
sm  0 

Now    L    sec    0  =  1;    hence,    since    the 

0=0 

a 

fraction   lies  between  two  values 

sin  0 

whose  limits  are  equal,  its  limit  must  be  the 


Fig.  5. 


same,  that  is, 


L  -^-  =    L 

0^osin0     0  =  0 


sin  0 


*  See  Holgate's  Geometry,  Arts.  356-361. 


20  FUNDAMENTAL   NOTIONS   AND   DEFINITIONS     [Chap.  I. 

Dividing  the  members  of  the  inequality  by  r  tan  0,  we  get 

cos  6  <  — —  <  1. 

tan  6 

Since  L  cos  6  =  1,  it  follows  that 

L  -±-=1  =    L  ^1. 

14.  Limit  of  a  monotone  function.  It  is  not  always  easy  to  find 
a  limit  of  a  function.  Moreover,  in  some  discussions  we  are  not 
so  much  concerned  as  to  what  the  limit  is  as  to  whether  the  given 
function  has  a  limit.  The  following  propositions,  which  we  give 
without  proof,*  will  aid  us  in  answering  this  question  for  certain 
types  of  functions. 

(a)  If  a  function  never  decreases,  but  always  remains  less  than 
some  constant,  then  it  has  a  limit. 

(b)  If  a  function  never  increases  but  always  remains  greater  than 
some  constant,  then  it  has  a  limit. 

Functions  of  the  kind  described  in  (a)  and  (b)  are  called  mono- 
tone functions. 

The  following  illustration  will  perhaps  aid  in  making  clear  the 
significance  of  these  statements. 

Given 

f(n)  =  0.333  .- 

=  _3 +JL  +  JL+  ...  +  A+  .... 

10     102     10*^       Tio* 

As  n  increases  f(n)  never  decreases,  but  always  remains  less  than 
0.4.  For  n  =  2,  f(n)  lies  between  0.3  and  0.4 ;  for  n  =  3,  between 
0.33  and  0.34;  for  n  =  4,  between  0.333  and  0.334,  etc.  Ifc  is  evi- 
dent that  as  n  increases,  the  range  of  values  within  which /(n)  must 
lie  constantly  decreases.  Consequently,  there  must  be  some  con- 
stant such  that  we  can  make/(n)  differ  from  it  by  as  little  as  we 
please  by  giving  n  an  integral  value  sufficiently  large.  We  know  in 
this  case  that  the  constant  in  question  is  ^,  but  that  is  not  essential 
in  establishing  the  fact  that  f{n)  has  a  limit  as  n  becomes  infinite. 

*  Geometrical  considerations  make  these  propositions  plausible.  The 
formal  proof,  while  not  difficult,  is  scarcely  within  the  scope  of  this  book. 
See  Pierpont's  Theory  of  Functions,  Art.  108. 


Arts.  14,  15]                   INDETERMINATE   FORMS 

EXERCISES 

Verify  the  following. 

1      L  x^-Sx  +  7  _25 

a?  =  3        X2  — 5               4 

0      z    (x-A)2-3/ix_1 
a=o      x(x  +  A) 

3.      L  sin x  +  y  cos x~y  =  sin  y. 

x  =  y           2                 2 

4.     Z  loS  (<*  +  *>=  log  a. 

5.      i  log^'2_6)=log2. 
a?=4           X+l 

6.     L  log  sin  0  =  0. 

2 

7.     £  [log  (x2  -  1)  - 

-log  (35-  I)]=l0g3. 

21 


15.    Indeterminate  forms.     It  may  happen  that  for  some  value 
of  the  variable  the  function  takes  an  illusory  or  indeterminate 

form,  as  -.     The  evaluation  of  such  an  indeterminate  form  will 

be  of  value  in  finding  the  derivatives  of  functions,  a  process  to  be 
considered  formally  in  the  next  chapter. 

x2  — 4 
Consider,    for   example,   the   function   y  = .     For   every 

x  —  2 

value   of    the   variable   x   other   than    x  =  2,   the   function   has 

a  definite  value,  but  for  x-—2   it   becomes      ~~     =-.     Strictly 

'  2-2     0  J 


speaking,  the  function 

hence,  in  order  to 
define  the  function 
completely  for  every 
value  of  the  variable, 
we  must  assign  it  a 
value  for  this  particu- 
lar value  of  the  vari- 
able. To  guide  us 
in  our  definition,  we 
make  use  of  a  simple 
graphical  representa- 
tion. For  x  different 
from  x  =  2,  the  equa 

ti0n  *-4 

y  =  — o 


X2-! 


has   no   definite   value   for   x  =  2; 


*•  X 


Fig.  6. 


22  FUNDAMENTAL   NOTIONS  AND   DEFINITIONS     [Chap.  I. 

reduces  to  y  =  x  -\-  2,  (1) 

whose  graph  is  the  line  EF,  Fig.  6.  Hence  for  values  of  x  other 
than  2,  the  values  of  the  function  are  represented  graphically  by 
points  on  the  line  EF,  as  B,  C,  D.  But  when  x  =  2  the  function 
has  no  definite  value,  and  hence  we  may  represent  it  by  any  point 
whatever  on  the  line  x  =  2,  that  is,  on  MJSf.  Of  the  values  which 
might  be  assigned  to  the  function  for  x  =  2,  there  is  one  value, 
represented  by  the  point  P,  which  is  the  limit  of  the  values  repre- 
sented by  the  points  on  EF&s  x  approaches  2  ;  and  it  is  convenient 
to  select  this  value  of  y  as  the  value  of  the  function  when  x=  2. 

We  define  therefore  the  value  of  the  function  for  the  critical 
value  of  the  variable  as  the  limit  that  the  function  approaches  as 
the  variable  approaches  the  value  for  which  the  function  becomes 

x2  —  4 

indeterminate.     Thus  for  x  =  2,  the  value  of  the  function  - 

x  —  2 

is  defined  as  L    — ^— . 

x~^x-2 

Any  definition  of  a  function  for  those  values  of  the  variable  for 
which  it  becomes  indeterminate  is  of  course  merely  a  convention; 
but  the  definition  adopted  in  this  case  is  a  very  useful  conven- 
tion, because  by  it  the  function  becomes  continuous  for  the  values 
in  question. 

In  general,  if  fix)  is  indeterminate  for  any  value  x  =  a,  the 

value  of  the  function  for  x  —  a  is  defined  by  the  limit      L  fix). 

x  =  a 
The  complete  definition  of  a  function  when  it  assumes  an  indeter- 
minate form  involves,  therefore,  the  determination  of  a  limit. 

Among  the  indeterminate  forms  that  a  function  may  assume 
are  the  following : 

5;      ».     oo-oo;     Oxx;     0°;     cc° ;      1*. 

0  GO 

An  example  of  the  first  form  is  given  by  the  function  — —  which 

x 

for  x  —  0  takes  the  form  -  .     The  value  of  this  function  for  x  =  0 
0 

is  therefore  defined  by  the  limit 

T    sii 

x  =  0    3 


L    smj=1  (Art.  13) 


Art.  15]  INDETERMINATE   FORMS  23 

The   function  x  log  x  for  x  =  0   has   the   form   Oxx,  and   the 

i 
function  (1  +  x)x  for  x  =  0  has  the  form  T°.     By  proper  trans- 
formations all  these  indeterminate  forms  may  be  reduced  to  the 

form  -. 

In  many  cases  the  limits  are  easily  found  by  obvious  algebraic 
transformations  or  by  the  use  of  series.  In  the  evaluation  of  the 
indeterminate  form  of  a  rational  function,  that  is,  the  quotient  of 
two  polynomials,  the  following  principles  may  be  used  to  advantage. 

1.  If  the  function  is  of  the  form  ^A_l  and  takes  the  form  -  for 

if/(x)  0 

x  =  0,  divide  both  numerator  and  denominator  by  the  lowest 
power-  of  x  that   occurs   in   either   numerator   or  denominator. 

If  the  fraction  becomes  —  for  x  =  oo  ,  divide  both  terms  of  the 

00 

fraction  by  the  highest  power  of  x  in  either  numerator  or 
denominator. 

2.  If  the  function  has  the  form  Q£l  and  takes  the  form  -  for 

$(x)  0 

x  =  a,  divide  both  <f>(x)  and  i[/(x)  by  the  highest  power  of  (x  —  a) 
common  to  both. 

The  student  should  note  that  we  do  not  divide  the  terms  of  the 
fraction  by  zero  or  infinity.  For  example,  while  the  division  by 
the  factor  (x  —  a)  holds  only  for  x  different  from  a,  the  values 
of  x  may,  however,  be  taken  as  close  as  we  please  to  a.  Hence, 
dividing  first  by  this  factor  and  afterwards  passing  to  the  limit, 
we  obtain  the  proper  result.  Indeterminate  forms  will  be  again 
considered  in  Chapter  XV. 

The  following  examples  illustrate  some  methods  that  may  be 
used  in  evaluating  various  indeterminate  forms. 

Ex.  1.     Find  the  value  of  a;2~6a; for  x  =  0. 

xs  —  4  x2  —  12  x 

x2-6x  L  .        x-6         _  1 


a5:i0iC3_4a.2_  12x      a.  =  ox2-4x- 12      2 

9  r2  -I-  S  a*3 
Ex.  2.     Evaluate  the  function  ^         for  x  =  co. 

x+5x3 

L    2 x2  +  3 xs  _    L     x _0  +  3_3 

+  5 


-t-523         03*00  1    ,  e      0  -f  5      5 


& 


24 


FUNDAMENTAL   NOTIONS  AND   DEFINITIONS     [Chap.  I. 


Ex.  3.     Find  the  value  of 


x3  -  3  x2  +  5  x  -  3 
2  x3  -  5  x2  +  8  x  -  5 


for  as 


x3-3x2  +  5x-3  =    (x2  -  2  a  +  3)  (a  -  1)    _=   a2  -  2  a  +  3 
2x3-5x2  +  8x-5      (2  a;2  -  2x  +  5)(x  -  1)      2x2-3x  +  5 


Hence,  we  have      i 


3  a2  +  5  x  -  3  __   L     x?  -  2  x  +  3 


=  i  2  x8  -  5  x2  +  8  x-  6 


1  2  x2  -  3  x  +  5      2 


Ex.  4.     Evaluate  a  ~  ^  ~  x'2  for  x  =  0. 


This  function  takes  the  form  - .  To  find  its  value  for  x  =  0,  we  multiply- 
both  terms  of  the  fraction  by  the  complementary  surd  a  +  Va2  —  x2 .  We 
then  have 

a  -  Vx2  -  x2  _     a2  -  (a2  -  x2)     _  x2 1      . 

x2  x2  (a  +  Va2  -  x2)      x2  (a  +  Va2  -  x2)      a  +  Va2  -  x2 ' 


and 


L 

x  =  0 


Va2 


<«-0a  +  Va2  —  x'1 


2a 


Ex.  5.     Evaluate  1  ~  cos  6  for  0  =  0. 


We  have 


Hence,  for  0  =  0, 


1 

2  sin2  - 
—  cos  0                2      0 

0                  0           2 

r  •    0' 
sin- 

2 
0 

2 

*=»2 

"  .    0" 
sin- 

2 

0 

2 

0=0 2 #«o 

"  .     0" 
sin- 
2 

0 

L    2 

-    2    J 

1  —  CO 

3*isd 

sflned  as  0. 

0. 


EXERCISES 
Find  the  limiting  values  of  the  following  functions  for  the  given  values  of 
the  variables. 


1. 


4  x3  -  3  x 


,  x  =  0. 


2  x8  —  3  x2  +  5  x 

3.    Vl  +  x  —  Vx,  x  =  co 

5.    *MU  =  4. 


2     x3  -  3  x2  -  4  x  +  12    x 


X2  +  4  ^  _  21 


Vi  +  x  -VT 


6.    ^-fl5,x  =  a. 


x  =  0. 


Art.  15]  INDETERMINATE   FORMS  25 


7.    Show  that  L  ^^  =  n. 


0  sin 


sin  nd         sin  n0       »  sin  nd 


sin  w0  w0  ,     n0  n0        1 

Suggestion  :  —. — -  =  n— — - ,     L  —. — -  = ■ — —  —  t 

yif  sin  0  sin  0  sin  0        T  sin  0       1 


8.    Show  that  for  0  =  0,  1  ~  C0S  -  =  0. 

sin  0 

10.   Show  that  L  isin|  =  L 

6  =  -  &=QX  4       & 

11.  Evaluate  —  +  -  ^  ~  5  -  for  x  =  0  and  x  =  oo  . 

3  a;4  -  2  x3  +  6  x 

12.  Evaluated   arctanx.  13.   Show  that  L  1  ~  cos  g  =  - . 

85  =  0  X  0  =  0  02  2 

MISCELLANEOUS   EXERCISES 

1.  Given  ?/  =  x2  —  3  as  +  5.  (a)  Find  the  increment  Ay  corresponding 
to  the  increment  Ax  of  the  variable.  (6)  Calculate  Ay  for  Xi  =  3  and  Ax  = 
0.1. 

2 .  Given  y  =  x8.     (a)  Find  the   expression  for  the  increment  Ay  for 

x  =  Xu     (b)  Find  the  limit  of  the  ratio  — ^  as  Ax  =  0. 

w  Ax 

3.  Given  y  =  sinx.  Take  X\  =  30°,  and  assume  for  Ax  the  following 
values:  1°,  30',  5'.     Make  a    table   showing   the   values   of   sin  (xi  +  Ax), 

A?/,  and— ^.     See  if  the  ratio  approaches  a  limit. 
Ax 

4.  If  <f>(x)  =  ax,  show  that  O(x)]2  =  0(2  x). 

5.  If   <f>(y)  =e*  +  e~y,  show  that  0(3  y)  =  OQ/)]3  -  3  <f>(y) 
and  0(x  +  y)  0(x  -  y)  =  0(2  x)  +  0(2 y). 

6.  Give  a  physical  illustration  of  a  function  that  approaches  a  limit  as 
the  variable  increases  indefinitely. 

7.  For  which  of  the  trigonometric  functions  is/(0)  =/(—  0)  ? 

8.  From  geometry  or  physics  give  two  or  more  examples  of  monotone 
functions. 

9.  Suppose  that  air  inclosed  in  a  cylinder  has  an  initial  volume  of  10 
cubic  feet  and  an  initial  pressure  of  20  lbs.  per  square  inch.  Assuming  that 
air  expands  according  to  Boyle's  law,  viz.  pv  =  constant,  calculate  the  value 

of  -£-  for  Au  =  2,  1,  0.1,  0.01.     Determine  the  same  ratio  for  Av  =  h,  and 
Av 

show  that  as  Av  =  0,  this  ratio  has  the  limit  —  2. 


26  FUNDAMENTAL  NOTIONS   AND   DEFINITIONS     [Chap.  I. 

10.  Show  that  L  (sec  x  —  tan  x)  =  0. 

2 

I 

11.  Examine  the  function  e*  for  continuity.  Draw  the  graph  of  the  func- 
tion from  x=—  2  to  x  =  +  2. 

12.  Find  the  limiting  values  of  the  following  functions  for  the  given  value 
of  the  variable : 

,  x   tan  6  —  sin  6    a      A                     /7  N   sec  6  —  1    a     n 
(«)  ^ .*  =  <>.  (6) ^—  ,0  =  0. 

13.  If  »  is  a  positive  integer  show  that 

x  =  a  x-  a 

14.  If  the  variables  in  any  given  functional  relation  are  interchanged 
(as  y  —  ex,  «  =  €»),  show  that  the  graphs  of  the  two  functions  are  symmetri- 
cal with  respect  to  the  line  x  —  y  =  0. 


CHAPTER   II 


DERIVATIVES  OF  ALGEBRAIC  FUNCTIONS 


16.  Increments.  The  idea  of  a  variable  as  employed  through- 
out the  calculus  is  but  little  used  in  elementary  mathematics. 
There  the  symbols  x,  y,  etc.,  stand  usually  for  unknown  but  fixed 
numbers  whose  values  are  to  be  determined.  Here,  as  in  the 
analytic  geometry,  we  associate  the  same  symbols  with  magnitudes 
that  change  or  grow  according  to  some  law  determined  by  the 
nature  of  the  problem  under  discussion.  As  we  have  already  seen, 
this  idea  of  growth  is  of  prime  importance  in  the  calculus,  and  it 
lies  at  the  basis  of  the  discussion  in  the  present  chapter. 

If  a  variable  be  assigned  some  arbitrary  value  as  xx,  a  function 
of  the  variable  takes  a  corresponding  value  f(x^) ;  and  if  the  vari- 
able changes  to  any  other  value  x,  the  function  takes  a  correspond- 
ing value  f(x).  The  changes  x  —  x1  and 
f(x)~f(xi)  of  the  variable  and  of  the 
function,  respectively,  are  called  incre- 
ments of  the  variable  and  of  the  function 
and  are  denoted  by  Ax  and  A/(x).  See 
Fig.  7.  We  have  therefore 
Ax  =  x  —  x1} 

A/0)  =/(*)  -/(*i) 

=f(x1  +  Ax)-f(xl). 
If  we  have  y  =f(x),  then  instead  of  writing  A/(#)  we  can  equally 
well  write  Ay. 

Given  the  increment  of  the  variable,  we  may  calculate  the  cor- 
responding increment  of  the  function,  as  shown  in  the  following 
examples. 


/(*)-/(*,) 


>X 


Fig.  7. 


Ex.  1.    Given  the  function 


Then 


/(x)  =  3x2  +  4x  +  2. 

=  3(xx  +  Ax)2  +  4(aci  +  Ax)  +  2 
=  6  cciAx  +  3 (Ax)'2  +  4  Ax. 
27 


(3  xi2  +  4  xi  +  2) 


28  DERIVATIVES   OE   ALGEBRAIC   FUNCTIONS     [Chap.  II. 

Ex.  2.  Given  the  law  of  falling  bodies  s  =  \  gt2.  Calculate  As  correspond- 
ing to  At  =  |,  h  =  4. 

We  have  here  As  =  \g  {t\  +  i)2  -  $  gh2 

Increments  are,  in  general,  variables  and  may  be  either  positive 
or  negative. 

As  was  shown  in  the  illustrative  problems,  Arts.  4  and  5,  the 
ratio  of  the  increment  of  the  function  to  that  of  the  variable 
is  useful  in  giving  the  slope  of  a  secant  line,  the  mean  speed 
of  a  moving  point,  etc.  In  fact,  this  ratio  of  the  increments 
always  gives  a  mean  value  of  some  kind.  It  is,  however,  the 
limit  of  this  ratio,  the  derivative,  that  is  of  special  importance. 

1  7.    Definition  of  a  derivative.     Consider  the  ratio 

Ay  =  Af(x)  =/(s1  +  AaQ--/(31> 
Ax         Ax  Ax 

xx  being  some  particular  value  of  x.  For  Ax  =  0  this  ratio  assumes 
the  indeterminate  form  -,  and  to  evaluate  it  for  this  value  of  Ax 

we  must  find  its  limit  as  Ax  =  0,  according  to  the  principles  laid 
down  for  the  evaluation  of  indeterminate  forms.  This  limit  is 
called  the  derivative  of  f(x)  with  respect  to  x  for  the  value  x  =  xv 
We  shall  denote  derivatives  with  respect  to  the  variables  x,  t,  etc., 
by  the  symbols  Dx)  Dt,  etc.     Thus  we  have  for  any  value  of  x, 

I>xf(a»=  L  /C*  +  *f)-fl«0=   L    **.  (1) 

Aa;±o  Ax  A«=oAa? 

The  derivative  may  therefore  be  defined  as  the  limit  of  the  ratio  of 
the  increment  of  the  function  to  that  of  the  variable  as  the  latter  in- 
crement approaches  the  value  zero. 

Instead  of  the  symbol  Dx,  other  symbols  are  often  employed. 
For  example,  having 

we   may   indicate   the   derivative  by  any  one  of  the  following 

symbols  : 

fix),  yj,  y',  Dy. 


Art.  17]  DEFINITION   OF   A   DERIVATIVE  29 

The  last  two  symbols  are  used  only  when  there  is  no  ambiguity 
as  to  the  independent  variable. 

The  derivative  is  also  referred  to  as  the  differential  coefficient  or 
as  the  derived  function,*  and  the  process  of  finding  the  derivative 
is  called  differentiation.  This  process  consists  of  the  following 
steps : 

1.  Give  to  the  independent  variable  an  increment. 

2.  Calculate  the  corresponding  increment  of  the  function. 

3.  Find  the  ratio  of  the  increment  of  the  function  to  that  of  the 
variable. 

4.  Determine  the  limit  of  this  ratio  as  the  increment  of  the 
variable  is  allowed  to  approach  zero. 

The  following  examples  illustrate  the  process  of  differentiation. 

Ex.  1.     Find  the  derivative  of  y  =f(x)  =  5  x2  —  3  x  +  4. 
First,  giving  the  variable  the  increment  Asc,  we  have 

y  +  Ay  =/(«  +  Ax)  =  5  (x  +  Ax)'2  -  B(x  +  Ax)  +  4 

=  5  x2  -  3  x  +  4  +  (10  x  -  3)  Ax  +  5  (Ax)2. 

Subtracting,  we  obtain  as  the  increment  of  the  function 

Ay=f(x  +  Ax)  -f(x)  =  (10  x-  3)  Ax  +  5  (Ax)2. 
The  result  of  dividing  by  Ax  is  the  ratio 

^  =  10x-3  +  5Ax; 

Ax 

and  the  limit  of  this  ratio  as  Ax  approaches  zero  is 

L    ^=     L     (10  x  -  3  +  5  Ax)  =  10  x  -  3. 

Ax  =  0  Ay        Aa'  =  p 

Hence,  /'(x)  =  D,  (5x2  -  3x  +  4)  =  10  x  -  3. 

To  find  the  value  of  the  derivative  for  some  particular  value  of  x,  we 
merely  substitute  that  value  of  x  in  the  derived  function  /'(x).  Thus,  for 
x  =  xi,  /'(xi)  =  10  xi  -  3  ;   for  x  =  5,  /'(5)  =  10  x  5  -  3  =  47. 


'  *The  word  "  derivative  "  is  frequently  used  to  mean  the  value  of  the  limit 
(1)  for  a  particular  value  of  x.  The  term  "derived  function"  refers  more 
properly  to  the  assemblage  of  all  such  values. 


30  DERIVATIVES   OF   ALGEBRAIC   FUNCTIONS     [Chap.  II. 

k 
Ex.  2.     Given  the  equation  expressing  Boyle's  law,  viz.  pv  =  k,  or  p  =  -  • 

Find  the  derivative  of  p  with  respect  to  v.  v 


We  have 

P  = 

k 

■  ~  •> 

V 

whence 

p  +  Ap  = 

k 

v  +  Av 

Subtracting, 

we 

get 

for  the  increment  of  p, 

Ap  = 

k 

v             v{\ 

Av 

v  -f  Av 

)  +  Av) 

whence 

Ap  _ 

Av 

:         k 

v(v 

1 

+  Av) 

and 

L   &  = 

Av  =  0  Av 

■■-k     L 

A»  = 

1 

o«(b  +  Av) 

v*' 

Hence  Dvp  =/'(«)  =-  \ 


v- 
For  k  =  200,  and  v  =  10  (see  Ex.  9,  p.  25), 

ft  (10)  =_200=_2. 

EXERCISES 

Find  the  derivatives  of  the  following  functions,  using  the  general  process 
described  in  this  section. 

1.   y  =  sc4.  2.    y  =  x'2  —  4  x  +  5. 

x2 
5.   y  =  xi-x-\-2.  6.   y  = 


x-1 

7.    y  =  {x-  a)s.  s.    s  =  \  gt*. 

9.    s  =  a«  +  ifir<2.  10.    P  =  ad  +  bd-h 

11.    H  p  (v  —  b)  =  fc,  find  Dvp. 

18.    Conditions   for   a   derivative.     Not   every   function   has   a 
derivative  for  all  values  of  the  variable.     In  order  that  the  ratio 
/fa  +  AaQ-Zfa)  ^ 

shall  have  a  limit  as  Aic  =  0,  it  is  necessary  that  the  numerator 
shall  approach  zero  simultaneously  with  the  denominator.  In 
other  words,  we  must  have 

L     /(«,  +  *»)  =/(*!), 
Az  =  0 


Arts.  18,  19]       SIGNIFICANCE   OF  THE   DERIVATIVE  31 

which  is  equivalent  to  writing 

L    f(x)=f(xx). 

X  =  X\ 

This  is,  however,  the  condition  for  the  continuity  of  f(x)  for  the 
value  x  =  xv  It  follows  that  a  function  cannot  have  a  derivative 
at  a  point  of  discontinuity. 

Although  the  numerator  of  (1)  must  vanish  simultaneously 
with  the  denominator,  it  may  happen  that  as  Ax  =  0  the  ratio 
becomes  infinite  for  particular  values  of  the  variable.  In  such 
cases,  we  say  that  for  the  particular  value  of  x  in  question,  say 
x  =  xx,  the  value  of  the  derivative  becomes  infinite,  and  write 

19.    Geometrical  and  physical  significance  of  the  derivative.     It 

A  f  (x) 
has  been  pointed  out  that  the  ratio     J  v  )  gives  the  slope  of  the 

Ax 

secant  line  passing  through  the  points  whose  ordinates  are 
yx  =f(x1),  y  =f(x1  +  Ax).  If  Ax  be  allowed  to  approach  zero,  the 
secant  line  approaches  as  a  limit  the  tangent  to  the  curve  y  =f(x) 
at  the  point  (x1}  yr). 

Hence,  if  y  =f(x)  is  represented  by  a  continuous  curve,  we  have 

f(x^  =  tan  <f>, 

where  <f>  is  the  angle  made  with  the  axis  of  X  by  the  tangent  to  the 
given  curve  at  the  point  (xl}  yx).  For  the  sake  of  definiteness  we 
shall  take  <f>  as  the  acute  angle  made  with  the  positive  direction 
of  the  X-axis.  It  may  be  either  positive  or  negative.  Tan  <£  is 
called  the  slope  or  gradient  of  the  tangent  to  the  curve;  hence 
the  derivative  of  the  function  f(x)  for  the  value  x  =  x1  represents  the 
slope  of  the  curve  y  —f(x)  at  the  point  (xl}  yj. 

Again,  it  has  been  shown  that  —  gives  the  mean  speed  of  the 

moving  point  during  the  time  A£.  If  A£  approaches  zero,  the 
limit,  that  is  the  value  of  the  derivative  Dts  for  t  =  tx,  gives 
the  speed  of  the  moving  particle  at  that  instant.  Similarly,  if  m 
denotes  the  mass  and  V  the  volume  of  a  body,  the  limit  of  the 

ratio  as  A V approaches  zero  gives  the  density  at  a  point;  and 


32  DERIVATIVES   OF  ALGEBRAIC   FUNCTIONS       [Chap.  II. 

if  AQ  denotes  the  heat  entering  a  body  and  At  the  corresponding 

rise  in  temperature,  the  limit  of  the  ratio  — — ,  as  At  approaches 

At 

zero,  gives  the  specific  heat  at  the  definite  temperature  rx. 

The  ratio  of  the  increments  gives  always  a  mean  value  for  an 
interval ;  the  derivative  an  instantaneous  value  or  the  value  at  a 
point. 

20.  General  theorems  on  differentiation.  While  the  process  of 
finding  a  derivative  described  in  Art.  16  is  sufficient  for  the  dif- 
ferentiation of  all  functions  that  can  be  differentiated,  it  becomes 
inconvenient  when  the  function  is  complicated,  and  the  labor  of 
differentiation  can  be  abridged  by  the  use  of  certain  general 
theorems  that  apply  to  all  classes  of  functions.  These  theorems 
are  given  in  the  following  Arts.  21-29.  Because  of  their  funda- 
mental importance,  the  student  should  have  them  at  his  ready 
command. 

21.  Derivative  of  a  constant. 


Let 

y  =  c. 

Then 

y  -f-  Ay  =  c. 

Subtracting,  we  have 

Ay  =  0, 

whence 

DxV=     L    *L  =  0. 

or 

Dxc  =  0. 

We  have  therefore  the  following  theorem  : 

Theorem  I.     The  derivative  of  a  constant  is  zero. 

Geometrically  the  equation  y  =  c  is  represented  by  a  straight 
line  parallel  to  the  X-axis.  The  slope  of  this  line  is  zero  in 
accordance  with  Theorem  I. 

22.    Derivative  of  a  variable  with  respect  to  itself. 

Given  y  =  x, 

then  y  +  A?/  =  x  +  A#, 

and  A?/  =  Ax. 


Arts.  20-24]  DERIVATIVE   OF  A   SUM  33 

Hence  Dxy  =     L    ^=     L    —  =  1; 

Ax^oAa,*     ^x  =  qAx 

that  is,  Dxa  =  1. 

This  result  gives  the  following  theorem : 

Theorem  II.     Tlie  derivative  of  a  variable  with  respect  to  itself 
is  unity. 

Ex.     Give  a  geometrical  illustration  of  Theorem  II. 

23.  Derivative  of  a  sum. 

Let  y  =f(x)  +  4>(x). 

Then  y  +  Ay  =f(x  +  Ax)  +  <f>(x  +  Aa>), 

and  Ay==/(a;  +  Aa;)— /(a;)   |  <ft(a?  +  Aa?)  -  <f>(x)  ^ 

Ax  Ax  Ax 

Taking  the  limits  of  both  members,  we  have 

L     ^/=     L    f(x  +  Ax)-f(x)        L   cf>(x  +  Ax)  -  <ft(a?) 
Ax  =  0Aa;    Ax^O  Ax  Ax  =  0  A# 

or  Dxy  =  DJ+Dx<f>. 

This  process  may  be  extended  to  the  sum  of  any  finite  number 
of  functions.     Thus,  if  we  have 

y  —  u  +  v  -f-  10, 

where  u,  v,  w  are  functions  of  x,  we  may  write 

Dxy  =  Bx(u  +  v  +  tu)  =  D^w  +  D^v  +  D^w. 

We  may,  therefore,  state  the  following  theorem : 

Theorem  III.     TJie  derivative  of  the  algebraic  sum  of  a  finite 
number  of  terms  is  the  algebraic  sum  of  their  derivatives. 

24.  Derivative  of  a  function  plus  a  constant. 

Let  y  =f(x)  +  c. 

By  the  application  of  Theorem  III,  we  have 
Dxy  =  DJ(x)  +  Dxc 

=  Dxf(x).  (Art.  21) 

That  is,  Dt[f(x)+c]  =  DJ(x). 


34 


DERIVATIVES   OF   ALGEBRAIC    FUNCTIONS     [Chap.  II. 


If  u  be  used  to  denote  f(x),  this  result  may  be  written 
2>a(tt  +  c)  =  Dxu. 

We  have,  therefore,  as  a  corollary  to  Theorem  III,  the  following : 

Corollary.  The  derivative 
of  a  function  is  not  affected  by 
increasing  or  decreasing  the  func- 
tion by  an  additive  constant. 

It  follows  also  that  two  func- 
tions differing  only  by  a  constant 
term  have  the  same  derivative. 
In  the  process  of  differentiating, 
the  constant  terms  may  conse- 
quently be  neglected, 
f  Geometrically,  this  corollary 
has  an  interesting  significance. 
Suppose  the  function  y  =f(x)  +  c  to  represent  some  curve.  The 
effect  of  adding  or  subtracting  a  constant  term,  that  is,  of  chang- 
ing the  value  of  c,  is  simply  to  shift  the  curve  up  or  down  with 
reference  to  the  X-axis.  (See  Fig.  8.)  As  has  been  shown,  the 
derivative  measures  the  slope  of  the  curve,  and  it  is  evident  that 
this  slope  for  any  particular  value  of  x,  as  a,  is  not  changed  by 
shifting  the  curve  as  indicated. 

This  corollary  has  also  an  important  significance  in  the  inverse 
operation  of  integration,  as  we  shall  see  later. 

25.    Derivative  of  a  product. 

Given  y  =  f(x)  •  <f>  (x). 

We  have  then       y  +  Ay  =  f(x  +  Ax)  •  <£  (x  +  Ax), 


Fig.  8. 


and 


Ay  =  f(x  +  Ax)  •  <j>(x  +  Ax)-f(x)  •  <f>(x) 

Ax  Ax 


By  adding  and  subtracting  f(x)  •  <f>  (x  +  Ax)  in  the  numerator, 
this  ratio  may  be  written  in  the  form 


Ax 


=  <f>(x  -\-  Ax) 


f(x  +  Ax)  -f(x)  4>(x  +  Ax)-<j>(x) 


Ax 


Ax 


Arts.  25,  26]  DERIVATIVE   OF   A   PRODUCT  35 

Since  <£(#)  is  by  hypothesis  continuous,  <f>(x-\-Ax)  in  the  limit 
becomes  <f>(x)  ;  hence,  we  have  upon  passing  to  the  limit, 

Dxy=<t>.Dxfi-f.Dx<l>. 

This  result  can  be  extended  to  the  product  of  a  finite  number 

of  functions ;  thus,  if 

y  =  uvw, 

where  u,  v,  w  are  functions  of  x,  we  have 

Dxy  =  X>a.  (uviv)  =  nv  •  Dxw  +  uw  ■  Dxv  +  vw  •  D^u. 

We  have  then  the  following  theorem : 

xftEOREM  IV.  TJie  derivative  of  the  product  of  a  finite  number 
of  factors  is  the  sum  of  the  products  obtained  by  multiplying  the 
derivative  of  each  factor  by  the  product  of  all  the  other  factors. 

26.   Derivative  of  a  constant  times  a  function. 

If  we  have  given  y  =  c-  f(%), 

we  have,  from  Theorem  IV, 

Dxy  =  cDxf(x)+f{x)-Dxc 
=  c  •  Dxf{x). 
That  is,  Dxcf(x)  =  c  .  Dxf(x). 

Again,  using  u  to  denote  f(x),  this  result  may  be  written 

t>x  (cii)  —  cDxu. 

Hence  we  have  the  following  corollary  to  Theorem  IV : 

Corollary.  The  derivative  of  a  constant  times  a  function  is  the 
constant  times  the  derivative  of  the  function. 

We  may  now  combine  the  results  of  Theorem  I,  the  corollary 
to  Theorem  III,  and  the  above  corollary  in  the  one  general  state- 
ment that  in  the  process  of  differentiation  constant  terms  disappear 
but  constant  factors  remain. 

Ex.     Find  the  derivative  of  y  =  x3  +  7  x  +  8. 

We  have  Dxy  =  Dx  (ar3)  +  Dx  (7  x)  +  Dx  8.  (Th.  Ill) 

Dx  (z3)  =  Dx  {xxx)  =  x2Dxx  +  x^D^  +  x*Dxx    (Th.  IV) 
=  3  x\  (Th.  II) 

Dx7x  =  7Dxx  (Cor.) 

=  7.  (Th.  II) 

Z>x8=0.  (Th.  I) 


36  DERIVATIVES   OF  ALGEBRAIC   FUNCTIONS     [Chap.  II. 

Combining  these  results,  we  have 

Dxy  =  3  x2  -f  7. 

The  student  need  not  write  down  each  step  as  has  been  done  in 
this  illustrative  example  ;  but  he  should  make  himself  so  familiar 
with  the  principles  set  forth  in  the  preceding  theorems  that  he 
can  write  down  the  results  at  once. 

EXERCISES 

Find  the  derivatives  of  the  following  functions. 

1.    y  =  x2.  2.    */=x2-10x  +  4.  3.    y=x2(x-l). 

4.   y  =  5x3-x(x  +  2).  5.    y  =  x2(x2  -  2)  +  x  (x  -f  3). 

6.  Plot  the  curve  y  =  4  x2  +  c,  giving  c  the  values  2,  4,  6,  8,  and  show  that 
each  of  these  curves  has  the  same  slope  for  x  =  2. 

7.  Let  the  distance  traversed  by  a  moving  point  be  given  as  a  function  of 
the  time  by  the  equation  s  —  80  t  —  16  t2.  Derive  a  general  expression  for  the 
speed  Dts,  and  find  the  speed  at  the  end  of  3  seconds. 

27.    Derivative  of  a  quotient. 

Let  y  =  m, 

then  y  +  A?/  =  ~ —     —4 , 

y-         *       e£(x  +  Ax)' 

and  consequently 

^      <£  (a;  +  Aaj)      <£  (x)  <f>  (x  +  Ax)  •  <£  (a?) 

Subtracting  and  adding  <f>  (x)  •  /(a?)  in  the  numerator,  we  have, 
after  dividing  by  Ax, 

.  /(x  +  Ax)-/(x)  _  _  <Ks  +  *»Q-*(«0 

A.V  _  Ax  ^  2__  AaJ 

Ax  <f>  (x  +  Ax)  •  <f>  (x) 

Kemembering  that   <f>  (x)  is   continuous   and   hence,  as   Ax  =  0, 
<£(x-f  Ax)  becomes  cf>(x),  we  have  for  the  limiting  value 

0  (x)-J./(s) -/(«)• -P.  »(«) 

"/"  [«#>(*)]2 


Arts.  27,  28]  DERIVATIVE   OF   A   QUOTIENT  37 

Denoting  the  two  functions  by  u  and  v,  respectively,  this  result 
can  be  more  conveniently  written  as  follows : 

V  JD™U  —  U  T>^V 


©- 


We  may  state  this  result  in  the  following  theorem : 

Theorem  V.  TJie  derivative  of  a  fraction  is  equal  to  the  denomi- 
nator times  the  derivative  of  the  numerator  minus  the  numerator 
times  the  derivative  of  the  denominator,  all  divided  by  the  square  of 
the  denominator. 

28.    Derivative  of  the  quotient  of  a  constant  by  a  variable. 

If  the  numerator  of  the  fraction  is  a  constant,  the  result  just 
obtained  becomes 


Hence  we  have  the  following  corollary  to  Theorem  V : 

Corollary.  The  derivative  of  the  quotient  of  a  constant  by  a 
variable  is  equal  to  minus  the  product  of  the  constant  and  the  deriva- 
tive of  the  variable  divided  by  the  square  of  the  variable. 

Ex.    Given  /(«)= ^^ ;  find /(a). 

JK  J     x2  +  3x-f  l'         J  K  J 

From  Theorem  V,  we  have 

fl(x)  =  (sa  +  3  x  +  1)  Dx(x  +  2)-(x  +  2)  Dx(x°-  +  3  x  +  1) 
J  K  J  (x2  +  3  x  +  l)2 

=  (a:2  +  3x  +  l)l-(x+  2)  (2  x  +  3)  =  _  (x2  +  4  x  +  5) 

(X2  +  3  x  +  1)2  (X2  +  3  x  +  1)2  * 

EXERCISES 

Find  the  derivatives  of  the  following  functions : 
1.    */  =  3x2-4x-6.  2.   ?/ =  4x3  -  2x2  +  5. 

3.   t/  =  x2(x-5).  4.   y  =  «(a;  +  1)(*  —  2). 

5.   y  =  x\x*  —  2)(*  +  l).  6.    ?/  =  x2(x-4)+x(x2  +  3). 

7.  ,-*-*'.  a  r-I. 

X2  X2 

9.    y  =      m     .  10.    y=ffa;+&. 

ax  +  6  ex  +  <Z 


38  DERIVATIVES   OF  ALGEBRAIC   FUNCTIONS     [Chap.  II. 


11.    P  =  0--.  12. 


P  t*  -  1 

>2_4x  +  5  14  as-6 


x2  +  2  *      x2  +  4  a;  -  6 

a:2 

15.  Plot  the  curve  */  =  —  +  c,  giving  c  the  values  1,  3,  5,  and  show  that 

each  of  the  curves  has  the  same  slope  for  x  =  3. 

16.  Show  that  if  the  nth  power  of  a  variable  occurs  as  a  factor  in  the 
denominator  of  a  fraction,  the  n  +  1st  power  of  the  factor  will  occur  in  the 
denominator  of  the  derivative  of  the  fraction  after  reduction  to  lowest  terms. 

29.  Derivative  of  un.  Let  u  denote  any  function  of  x.  To 
determine  the  derivative  of  un  with  respect  to  x,  we  proceed  in 
the  usual  manner.     Thus,  let 

V  =  u1\ 
then  y  +  Ay=(u  +  Au)n, 

and  provided  Au  =£.  0,  we  have 

Ay  _  (u  +  Au)n  —  un      Au  -s 

Ax  Au  Ax 

Since  u  is  continuous  and  therefore  Au  approaches  zero  with  Ax, 
we  have  upon  passing  to  the  limit 

Dj,=     L  £*.     L    (»  +  *")'-«?  .     L    *2.  (2) 

Az  =  oAa     Am  =  0  Am  Aa=oAa 

The  evaluation  of  the  first  limit  in  the  second  member  requires 
consideration  of  the  special  limit 

L  vn-an 

x  =  a    <K  -  a 

Let  n  be  any  positive  integer  ;  then  by  division  we  obtain 
xn  —  a* 


x—  a 
whence 

xn  —  an 


xn-\  _|_  xn~2a  +  •••  +  xan~2  +  an~\ 


In  the  parenthesis  there  are  n  terms,  each  of  which  has  the  limit  an-1  as 
x  =  a.  Since  n  is  finite,  the  limit  of  the  sum  is  equal  to  the  sum  of  the  limits, 
and  we  have  therefore 

i  t«!=M«-i  (3) 

x  =  a  x—  a 


Arts.  29,  30]  DERIVATIVE   OF   IP  39 

It  is  easily  shown  that  (3)  holds  when  n  is  a  positive  fraction,  also  when 

n  is  negative,  either  integral  or  fractional.     For  the  first  case  let  n  =-£  and 

Q 
assume  x  =  *«,  a  =  &«  ;  for  the  second,  put  n  =  —  m.     The  proof  is  similar  to 
that  just  given. 

In  (3)  let  x  be  replaced  by  u  -\-  Au  and  a  by  u  j  then  as  a;  =  a, 

Am  =  0,  and  (3)  becomes  L   0  +  A^)"-^w  =  nw-i  /4) 

Aw  =  0  A« 

Using  this  result  in  (2),  we  obtain  when  n  is  a  positive  integer 

Dx(un)=nu"-1Dxu.  (5) 

In  Art.  65  it  will  be  shown  that  formula  (5)  holds  good  when 
n  is  any  real  constant,  rational  or  irrational. 

If  u  =  x,  we  have,  since  Dxu  =  1,  the  important  special  case 

^")  =  «"'  (6) 

Ex.  1.    Let  y  =  o(8  z2  +  7)3  ;  find  Dxy. 

Put  m  =  3  x2  -f  7. 

We  have  then  Dxy  =  a  ■  Dxu'^  =  3  au2Dxii. 

But  Dxu  =  DX(S  x*  +  7)  =  6  x. 

Combining  these  results,  we  obtain 

Dxy  =  18  ax(3  x2  +  7)2. 

Ex.  2.   Let  y  =  Vx2  —  a2. 

Substituting,  u  =  x2  —  a'2, 

we  have  y  =  vr. 

_i  i  ~ 

Hence  2)x?/  =  |  u  *  Dxu  —  \u  ^  2x  = 


y/x2  -  a2 

30.   Explicit  and  Implicit  Functions.     Algebraic  functions.     If  a 

function  is  expressed  directly  in  terms  of  the  variable,  it  is  called 
an  explicit  function  of  that  variable ;  if,  on  the  other  hand,  it  is  not 
expressed  directly  in  terms  of  the  variable,  it  is  called  an  implicit 
function.     Thus, 

y  =  3  x2  -f  7  x  +  9 

is  an  explicit  function  of  x,  while  in 

3xy  +  7y-9x2  =  4; 


40  DERIVATIVES   OF   ALGEBRAIC   FUNCTIONS     [Chap.  II. 

y  is  an  implicit  function  of  x.  We  can  express  y  as  an  explicit 
function  of  x  by  solving  the  above  equation  for  y  and  thus  obtain 

4  +  9Z2 

v=  — 

u      Sx  +  1 

A  function  of  the  general  form 

y  =  a0xn  +  aYxn-1  H f-  an 

belongs  to  a  class  of  functions  known  as  algebraic  functions.1*  In 
the  form  in  which  the  function  is  here  written  it  is  also  an  ex- 
plicit function  of  x.  The  general  laws  for  differentiation  that 
have  been  developed  in  the  preceding  articles  are  sufficient  to 
enable  the  student  to  write  at  once  the  derivative  of  any  explicit 
algebraic  function.  The  differentiation  of  implicit  functions  will 
be  considered  later. 

Ex.    Differentiate  the  function  y  =  x3  —  4  x  +  2  x~l. 
From  Art.  23,        Dxy  =  Dx(x*)  -  Dx(i  x)+  I)x(2  ar1), 
whence,  using  (6) ,  Art.  29,  we  have 

Dxy  =  3  x2  -4  -  2ar*. 

EXERCISES 

Differentiate  the  following  algebraic  functions. 

1.  y=x?.  2.  y  =  x2  -  x-1  +  5. 

3.  y  =  x^-x~K  4.  y=(2x-by. 

5.  y  =  (x2  -  4  x  +  3)2.  6.  y  =  (x2  -  arf. 

7.  y  =  ^lT^2.  8.  r=(4«  +  8)7*. 

Q  „ lx'2  —  a?  ,n  .       a  .    b   .    c 

U,'"il  +  7+«'  12.  (>  =  «•  + M*- 2. 

13.    s  =  80-16*2.  14-   If  =*»(!-»)•. 

i 


15.    2/  =  x8\/x2-5.  16.    y  =  x*(x2  -  a2)  », 

17.  By  differentiating  — ,  show  that  Dxxn  =  nxn~l  holds  for  negative  inte- 
gral values  of  n.  x 

_y 

18.  Given  xe  3  =  c ;  express  ?/  as  an  explicit  function  of  x. 


*  For  a  more  general  definition  of  an  algebraic  function,  see  First  Course, 
p.  9. 


Arts.  30,  31]  FUNCTION   OF  A   FUNCTION  41 

19.    Given  x  =f(<f>)  =  a(<f>  —  sin  <f>) 

y  =  F(<t>)  =  a(l  —  cos  0), 
express  &  as  an  explicit  function  of  y. 

31.  Derivative  of  a  function  of  a  function.  The  derivative  of  un 
discussed  in  Art.  29,  where  u  is  a  function  of  x,  is  a  special  form 
of  the  more  general  case  which  we  shall  now  consider.  Let  us 
suppose  that  y=f(u),  and  u  =  <f>(x),  where  these  functions  have 
respectively  the  derivatives  f'(u)  an(i  4>'(x)  f°r  tne  corresponding 
values  of  the  variables  u  and  x.  If  we  wish  to  find  Dxy,  we  can 
do  so  by  substituting  the  given  value  of  u  mf(u),  thus  expressing 
y  directly  as  a  function  of  x ;  from  this  expression  we  can  obtain 
the  desired  result  by  methods  already  explained.  In  many  cases, 
however,  the  derivative  can  be  obtained  more  easily  by  the  follow- 
ing method. 

Let  x  take  an  increment ;  then  u  and  y  will  also  take  increments, 
and  since  u  and  y  are  continuous  functions  of  xf  these  increments 
will  approach  zero  as  the  increment  of  x  approaches  zero.  We 
have  then 

Ay=/(te  +  A«)— /(«). 

Ay  =f(u  +  Am)  -  /(it)  t  A^  Aw^tO. 

Ax  Au  Ax 

Since  Au  =  0  as  Ax  =  0,  we  have  in  the  limit 

L    *?=     L    /("  +  Am)-/00     x    A». 
Aa;i0Aa?      Ait  =  0  A?*  Ax  =  oA#' 

that  is,  2>a.2/  =  DUV  ■  T>^u.  (1) 

This  formula  expressed  in  words  gives  us  the  following  theorem : 

Theorem.  Ify  =  f(u)  and  u  =  <f>(x),  the  derivative  of  y  with 
respect  to  x  is  the  product  of  the  derivative  of  y  with  respect  to  u  and 
the  derivative  ofu  ivith  respect  to  x. 

This  theorem  asserts  the  principle  that  if  u  changes  m  times 
as  rapidly  as  x,  and  y  changes  n  times  as  rapidly  as  u,  then 
y  changes  mn  times  as  rapidly  as  x.  For  example,  if  a  horse 
travels  twice  as  fast  as  a  man,  and  a  train  four  times  as  fast  as 
the  horse,  the  train  travels  eight  times  as  fast  as  the  man. 


42  DERIVATIVES   OF  ALGEBRAIC   FUNCTIONS     [Chap.  II 

Ex.  1.  Find  Dxy  when  y  =  Vu  and  u  =  3  x2  —  4. 

We  have  Duy  =  £  w~2,  and  Z>xw  =  6  x ; 

hence  #**/  =4m"^6x  =  3  seiT^  = — 


V3  x2  -  4 

_3 

Ex.  2.    Find  Z^  when  ?/  =  (a2  —  x2)  *. 

-3-  _5 

Put  tt  =  a2  —  x2 ;  then  y  =  u  2,  Z>M?/  =  —  § u  2,  and  Dxu  =  —  2  x. 
Therefore  A#  =  3  xu~%  = — 


(a2  -  x2)1- 


EXERCISES 

Differentiate  the  following  functions  : 

1.   y  =  ^x2  +  5.  2    y=(x2-2x+5)i 

*»-<*-*><■  4.  ,=(«*.*)-* 

5.  y  =  (x*  +  a3)2.  i 

, y  /      x4      \  3 

7.    y  =  #8a*-4aa.  .  6.   y  =  ^^_^J   . 

9.    y  =  (x*-2x  +  6)"*.  B    y  =  V(x  +  m)(,  +  ,}> 

13-    P  =  \Y+J2'  ia    2/=3(x2-4)^(x  +  3)i 

13.    y  =  (x+Vx2-a2)i  12.   p=(l  +  **)"*. 

15.   v=       *       .  14.    —    *2  +  2 


V  a2  —  x2  Vx2  +  1 

16.  By  putting  f(u)  =  cu,  prove  the  corollary  to  Theorem  IV,  Art.  26. 

17.  By  putting  /(w)  =  wM,  develop  the  law  for  differentiation  of  un  given 
in  Art.  29. 

18.  Write  a  function  which  has  x2  as  its  derivative. 

19.  Show  that  the  curve  for  which  the  slope  of  the  tangent  line  (that  is, 
tan  </>)  is  numerically  equal  to  the  abscissa  has  the  general  form  y  =  \  x2  -f  C. 

20.  Find  the  values  of  x  for  which  the  derivative  of  /(x)  =  x3  —  9  x2  +  24  x 
becomes  zero.     What  is  the  geometrical  interpretation  ? 

21.  At  what  points  of  the  curve  y  =  x8  —  12  x  is  the  tangent  parallel  to 
the  X-axis  ? 

32.  Derivatives  of  inverse  functions.  If  y  is  given  as  an 
explicit  function  of  x,  and  it  is  possible  to  solve  for  x  in  terms 
of  y,  we  may  then  express  x  as  an  explicit  function  of  y. 
Thus,  if 


Art.  32]  INVERSE   FUNCTIONS  43 


y  =  ar 


we  have  x=  ±  Vy  +  4. 

In  general,  let  x  =  <f>(y)  be  the  result  of  solving  the  equation 
y=f(x)  for  »;  then  f(x)  and  £(y)  are  said  to  be  inverse  functions. 
Examples  of  inverse  functions  are  sin  x  and  arc  sin  y,  log  x  and  e\ 

\_ 
xn  and  yn,  etc. 

In  the  process  of  differentiation  we  ordinarily  express  y  as  an 
explicit  function  of  x  and  determine  Dxy  directly.  Sometimes, 
however,  it  is  convenient  to  express  a?  as  a  function  of  y  and  find 
Dxy  by  means  of  the  inverse  function.  To  find  the  relation  between 
the  two  derivatives,  we  may  proceed  as  follows : 

Given  y=f(x);  (1) 

and  let  the  inverse  function  be 

»  =  *(y).  (2) 

By  differentiation,  we  have  from  (2)  by  aid  of  Art.  31, 

l  =  D„<j>(y).Dxy, 
or  1  =  Dyx  •  Dxy. 

Hence  we  have,  provided  Dyx  ^  0, 

This  formula  states  the  principle  that  if  y  is  changing  n  times 
as  rapidly  as  x,  then  x  is  changing  -th  as  rapidly  as  y. 

Ex.  Given  the  equation  of  the  parabola  y2  =  4  px  ;  find  Dxy.  Solving  for 
x,  we  have 

y2 
4p 

■^         2  v        V 
whence  Dyx  =  -r1 -  =  ~  • 


From  the  theorem  of  inverse  functions,  we  have 
Dxy 


1     _2p  _     2p__jpt 

~~^Z        \  nr. ' 


Dyx       y        -y/4  px 


44  DERIVATIVES   OF   ALGEBRAIC   FUNCTIONS     [Chap.  II. 

EXERCISES 

1.    Find  Dxy  when  y2  =  x  -  4.  2.    Find  Dxy  when  y  —  =  x2. 

2/+  1 

3.    If  p2  =  a0  +  6,  find  Z>0p.  4.    If  v2  =  2  gs,  find  D.t;. 

5.    Find  functions  inverse  to  the  following  functions  : 


(a)  y  =  xs;     (6)  y  =  log  (x2  +  a2)  ;     (c)  y  =  Va2  -  z*. 
6.    Find  the  function  inverse  to 


/(z)  =  \og(x+Vx2  +  l). 

33.   Derivative  of  one  function  with  respect  to  another,  when  both 
are  functions  of  a  common  variable. 

Suppose  we  have  given 

y  =  f(t),        x  =  <f>(t), 
and  that  we  wish  to  find  Dxy.     We  have,  for  At  =£  0, 

Ay 
Ay  _  At 
Ax      Ax 
At 
Because  of  the  continuity  of  f(t)  and  <f>(t),  Ay  =  0  and  Aa?  =  0 
simultaneously  as  At  ==  0.      Hence,  by  Art.   17,   we  have   upon 
passing  to  the  limit. 

Ex.  1.     Given  y  =  z2  +  1,  and  x  =  y/z ;  find  Dxy. 
We  have  Dzy  =  2  *, 

and  D^  =  £  2_2\ 

Hence,  X>xy  =  -^-  =  4z^4x3. 

Ex.  2.     The  equations  s  =  ^  #£2  and  v  =  gt  refer  to  falling  bodies.     Find 
the  derivative  Z).,v. 

2>,*  =  22  =  £  =  I  =  2. 

As      gt     t      v 

EXERCISES 

1.  Find  2>xy,  when  ?/  =  3  £2  —  t  -  10  and  z '■  =  t  +  8. 

2.  Find  Dxy,  when  ?/  =  t*  —  3  £2  +  7  and  re  =  £2  —  2  £  +  4.     Find  the  values 
of  this  derivative  for  t  =  0,  2,  5. 


Art.  33]  FUNCTIONS   OF  A   COMMON   VARIABLE  45 

3.  IiP=  Vtand  6  =  t2  -  10,  find  D6p. 

4.  If  x  =  at  and  y  =  bt  —  |  c£2,  find  Dxy  and  Dyx. 

5.  Work  examples  3  and  4  by  eliminating  t  before  differentiating. 

6.  Prove  the  theorem  of  Art.  33  by  making  use  of  the  theorems  of  Arts. 
31  and  32. 

MISCELLANEOUS   EXERCISES 

Differentiate  the  following  functions  with  respect  to  the  variable  indicated. 

1.    ?/  =  5x4  +  3.r3-8x  +  7.  2.    y  =  x2  (3  x2  -  4)(x  +  1). 

ill  q      a  v 

3.   y  =  x*  (x*  -  a*)2.  4.   p=    ' 


2  (x  -  2)2 

(L: 
7.   */  =  |(Vx  +  a)2  -a(Vx-t-a)^. 


s.  y  =  ^-2>vr+?.  6.  y  =  a  +  »^^-v 


8 

v-      *M 

(1  +  X)» 

10. 
12. 

y  =  (x  +  Vx2  —  1)H. 
iz  +  1 

14. 

.._  (l-z)Vl-x2 

»      * 


x"(a  +  x)™ 


11.  */  =  Vx  +  Vl  +  x2. 
3  ' 


13.   2/  =  A/  — 

*  x  — 


15.     y=    VaP  +  l+vte-l 

+  »  '  Vx^+l  -  vx2^-! 

1 


16.    y  =  —  ^2  ~  a^ 3  .  1.1.    y  = 


x  +  a  x  +  Vx2 

18.   v  =  2g±J— .*.  is.  p=     «     -. 


VX2  +  1  +  X 

20.  Given  t>  =  - m  (1  +  ap),  where  m,  (7,  and  a  are  constants,  find  2>»p. 

21.  Show  that  the  slope  of  the  tangent  to  the  curve  y  =  x3  +  4  is  never 
negative.  Find  the  slope  for  x  =  0,  x  =  2.  For  what  values  of  x  does  the 
slope  decrease  as  x  increases  ? 

22.  Find  the  angle  which  the  tangent  to  the  parabola  y2  =  9  x  at  the  point 
(4,  6)  makes  with  the  X-axis. 

23.  Given  y  =     u  +  5      and  u  =  x2  -  5  ;  find  Dxy. 

Vw2  -  15 

24.  Given  s  =  bt  +  |  a£2  and  v  =  at,  find  Dsv  in  terms  of  v. 

25.  The  equation  pv  —  C  expresses  Boyle's  law,  C  being  a  constant.  Find 
D0p  and  Z?/>v. 


46  DERIVATIVES   OF  ALGEBRAIC   FUNCTIONS     [Chap.  II. 

26.  From  Regnault's  experiments,  the  heat  q  required  to  raise  the  tem- 
perature of  a  unit  weight  of  water  from  0°  C.  to  a  temperature  t  is  given 
by  the  equation 

q  =  r  +  0.00002  r2  +  0.0000003  r3. 

(a)  Find  DTq.     (b)  Calculate  the  numerical  value  of  DTq  for  t  =  35°. 

27.  The  efficiency  of  a  screw  as  a  mechanical  device  is  given  by  the 
relation  E  =  -  *    —  ^x> ,  where  /x  denotes  the  coefficient  of  friction  and  x 

the  tangent  of  the  pitch  angle  of  the  screw.    Find  the  derivative  DXE,  and  its 
values  for  x  =  0  and  x  =  /j..     Find  also  the  value  of  x  for  which  DXE  =  0. 

28.  Find  the  slope  of  the  tangent  to  the  circle  x2  +  y2  =  25  for  the  points 

x  =  0,  x  =  4,  x  =  —  3. 


29.  Find  Dsv,  where  v2  =  2  c  ( )• 

\8      a) 

30.  Dxy  =      x2      and  u  =  -  x2  4-  &x.     Find  Duy. 

ax  +  b  2 

31.  Interpret  geometrically  the  corollary  of  Art.  26. 

32.  Find  functions  which  have  the  following  derivatives  : 

(a)  3x2;        (5)  4  x*  -  1 ;         (c)  x2  -  x  +  2  ;         (<j)  -  I . 

x'2 

33.  Deduce  Theorem  V,  Art.  27,  from  Theorem  IV  by  means  of  the  sub- 
stitution -  =  w,  whence  l)xu  =  Dx(vw). 

34.  If  the  factor  (x  —  a)  occurs  n  times  in  a  function  /(*),  show  that  it 
occurs  n  —  1  times  in  the  derivative  Dxf(x),  and  from  this  principle  deduce  a 
method  of  finding  whether  an  algebraic  equation  has  multiple  roots. 


CHAPTER   III 

ELEMENTARY   APPLICATIONS   OF   DERIVATIVES 

34.   Slope  of  a  curve.     The  equations  of  the  curves  thus  far 
discussed  have  all  been  given  in  the  form 

y  =/(?),  ovF(x,y)  =  0.  (1) 

In  some  cases  it  is  more  convenient  to  express  both  x  and  y  in 
terms  of  a  third  variable ;  thus 

x  =  <j>(t),  y  =  *Kt),  (2) 

where  <£(£),  \f/(f)  are  single-valued  functions  of  the  variable  t.  We 
call  these  equations  the  parametric  equations  of  the  curve.  As 
examples  of  parametric  equations  of  curves,  we  have  for  the 
circle  the  equations 

y  =  a  sin  6,  x  =  a  cos  0, 
and  for  the  cycloid 

y  =  a  (1  —  cos  6), 

x  =  a(0  —  sin  6). 

We  may  pass  from  the  parametric  form  of  expression  to  that 
given  by  equation  (1)  by  eliminating  the  common  variable. 

In  whichever  form  the  equation  of  the  curve  is  given,  we  have 

2)^  =  tan*,  (3) 

where  <j>  denotes  the  acute  angle  between  the  tangent  to  the  curve 
and  the  positive  direction  of  the  X-axis.  To  determine  tan  <j>  from 
the  parametric  equations  of  the  curve,  we  have  by  Art.  33 

A2/  =  g  =  tan*.  (4) 

Tan  cf>  has  been  defined  as  the  slope  or  gradient  of  the  tangent  to 
the  curve.  It  may  also  be  called  the  slope  of  the  curve,  for  the 
direction  of  the  tangent  to  the  curve  is  also  the  direction  of  the 

47 


48       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 


curve  at  the  point  of  tangency.     The  slope  of  the  curve,  therefore, 
is  given  by  the  value  of  the  derivative  Dxy. 

The  value  of  the  derivative  at  any  point  gives  still  other 
properties  of  the  curve.  Since  the  direction  of  the  tangent  is 
identical  with  the  direction  of  the  curve  at  the  point  of  tangency, 
it  follows  that  so  long  as  <f>  is  a  positive  angle  the  ordinates  of 
the  curve  are  increasing  as  the  abscissa  increases.  When,  how- 
ever, <£  is  positive,  tan  <£,  and  hence  Dxy,  is  positive ;  similarly, 
when  <j>  is  negative,  the  ordinates  of  the  curve  are  decreasing  as 
the  abscissa  increases.  But  in  this  case  tan  <£,  and  hence  Dxy,  is 
negative.  If  the  value  of  Dxy  becomes  zero  for  any  value  of  x, 
then  tan  <f>  is  zero  and  cf>  must  be  zero ;  that  is,  the  tangent  to  the 
curve  at  this  point  is  parallel  to  the  X-axis.  This  occurs  when- 
ever the  curve  has  a  turning  point,  that  is,  a  point  where  the  or- 
dinates cease  to  increase  and  begin  to  decrease,  or  vice  versa. 
Turning  points  are  shown  at  A,  C,  and  E,  Fig.  10.  If  Dxy  be- 
comes infinite  for  a  particular  value  of  x,  the  value  of  <£  is  then 
\  it  or  —  \  it  ;  that  is,  the  tangent  to  the  curve  at  that  point  is  per- 
pendicular to  the  X-axis.  We  may  summarize  these  results  as 
follows : 

At  any  point  of  the  curve  y  =f(x),  the  ordinate  increases  or  de- 
creases with  increasing  x  according  as  Dxy  is  positive  or  negative. 
If  Dxy  is  zero  for  any  value  ofx,  then  at  that  point  the  tangent  to  the 
curve  is  parallel  to  the  axis  of  abscissas.  If  Dxy  becomes  infinite  for 
any  value  of  x,  the  tangent  to  the  curve  is  perpendicular  to  the  axis 
of  abscissas  at  that  point. 

Ex.  1.  Investigate  the  curve  y  =  ^  x3  —  x 
by  means  of  its  derivative. 
Differentiating,  we  get 

=  8£ _  j  =  (x  +  2)(*-2) 
xJ        12  4 

For  x  >  2  or  cc  <  —  2,  Dxy  is  positive, 
and  consequently  y  increases  with  x. 
For  values  of  x  between  2  and  —  2,  Dxy  is 
negative,  and  y  therefore  decreases  as  x 
increases.  For  x  =  2  and  for  x  =  —  2, 
Dxy  =  0  ;  hence  at  these  points  the  tangent 
to  the  curve  is  parallel  to  the  X-axis.  The 
curve  is  shown  in  Fig.  9. 


Fia.  9. 


Arts.  34,  35]  DERIVED   CURVES  49 

EXERCISES 

Investigate  the  following  curves  by  means  of  their  derivatives. 

1.    y  =  xs-Sx2  +t>ie.  2.   y2  =  8x-10.  3.   y=  — . 

X 

4.  For  what  values  of  x  does  the  function  x2  +  -  increase  with  x  ?    For 

x 
what  value  does  it  decrease  as  x  increases  ? 

5.  The  equation 

t  =  53.6 p* -35.7 

gives  approximately  the  temperature  r  of  steam  as  a  function  of  its  pressure 
p.  (Here  r  is  in  degrees  Fahrenheit,  and  j?  in  pounds  per  square  foot.) 
Show  how  the  derivative  Dpt  changes  as  the  pressure  increases,  and  sketch 
roughly  the  curve  r  =f(p). 

6.  A  cylindrical  vessel  with  one  end  open  is  to  hold  300  cu.  in.  The 
superficial  area  A  of  this  vessel  (cylindrical  wall  plus  one  base)  is  a  function 
of  the  radius  of  the  base.  Deduce  the  equation  and  examine  it  by  means  of 
the  derivative.     Interpret  the  results. 

7.  The  efficiency  rj  of  a  hoisting  device  is  a  function  of  the  load  P  raised 
as  expressed  by  the  equation 

P 
mP  +  n 

where  m  and  n  are  constants.  Show  how  the  derivative  varies  with  the  load 
P  and  sketch  the  general  form  of  the  curve  rj  =/(P). 

8.  By  means  of  the  derivative  investigate  the  curves : 

(a)  y  =  a  +  b{£-  c)»  ; 
(6)  y  =  a  +  b(x  -  c)7. 
Show  the  form  of  the  curves  in  the  vicinity  of  the  point  (c,  a). 

9.  Given  the  continuous  curve  y  =/(:»).  Show  by  means  of  the  graph 
that,  for  positive  values  of  Ax,  Ay  is  positive  or  negative  according  as  the 
function  is  increasing  or  decreasing  at  a  given  point.    By  considering  the 

limit  L  — ^  deduce  the  general  law  given  on  p.  48. 
Aa?=oAx 

35.  Derived  curves.  The  derivative  of  a  function  is  also,  in 
general,  a  function  of  the  independent  variable,  and  may  be  rep- 
resented by  a  graph  in  the  same  manner  and  under  the  same 
conditions  as  the  original  function.  This  curve,  whose  equation 
is  y=f'(x),  is  known  as  the  derived  curve. 


50       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 

The  principles  developed  in  the  last  article  enable  us  to  estab- 
lish certain  general  properties  of  derived  curves.  Thus  the  graph 
of  y  =f'(x)  crosses  the  axis  of  x  at  those  points  for  which  the 
original  curve  has  a  turning  point.  Moreover,  we  can  tell  whether 
the  derived  curve  -crosses  the  X-axis  from  above  to  below  or  vice 
versa  ;  for  we  know  that  as  x  increases,  f'(x)  is  positive  or  nega- 
tive according  as  f(x)  increases  or  decreases.  Hence,  if  the  values 
of  f(x)  are  increasing  as  the  turning  point  is  approached,  and  de- 
creasing after  it  is  passed,  as  at  points  A  and  E,  Fig.  10,  the 


r 

p^~\ 

E 

A  V 

I 

\yT 

B 

7    > 

Si 

A 

c 

JE« 

\ 

K 


Fig.  10. 


values  of  fix)  pass  from  positive  to  negative,  and  the  derived 
curve  passes  from  the  upper  to  the  lower  side  of  the  axis,  as 
shown  at  points  A'  and  E'.  At  the  turning  point  C,  on  the  other 
hand,  f(x)  changes  from  a  decreasing  to  an  increasing  function 
and  f'(x)  passes  from  negative  to  positive  values ;  that  is,  the 
derived  curve  passes  from  the  lower  to  the  upper  side  of  the 
X-axis  at  the  point  C. 

Where  the  derived  curve  has  turning  points,  as  at  B'  and  D\ 
the  curve  y  =  f(x)  has  points  of  inflection,  as  B  and  D.  See 
Art.  88. 

For  those  functions  that  are  to  be  considered  in  the  present 
volume  the  derived  curve  is,  in  general,  a  continuous  curve.  We 
shall  meet,  however,  two  exceptional  cases  in  which  it  becomes 
discontinuous. 


Art.  35] 


DERIVED   CURVES 


51 


1.  When  the  tangent  to  the  curve  yz=f(x)  becomes  perpen- 
dicular to  the  X-axis,  as  at  M,  Fig.  11.  Here  the  value  of  /'(a?) 
becomes  infinite  and  the  derived  curve  m  has  infinite  branches. 

2.  When  the  curve  y  =  f(x)  has  an  angular  point,  as  at  N, 
Fig.  12.  At  such  a  point,  the  limiting  position  of  the  tangent  as 
the  point  of  tangency  approaches  N  from  the  left  is  different 


Fig.  11. 


Fig.  12. 


from  the  limiting  position  as  the  point  of  tangency  approaches 
N  from  the  right.  Hence,  the  value  of  f'(x)  takes  a  sudden 
jump,  and  the  derived  curve  n  has  a  discontinuity. 

The  ordinates  of  the  derived  curve  are  the  successive  values  of 
Dxy  of  the  original  curve.  Since  Dxy  measures  the  slope  of  the 
tangent  to  the  original  curve,  it  follows  that  for  any  value  of  x 
the  ordinate  of  the  derived  curve  measures  the  slope  of  the 
original  curve. 

It  is  evident  that  the  slope  of  the  curve  y=f(x)  is  independent 
of  the  position  of  the  axes  so  long  as  they  remain  parallel  to  their 
original  position.  We  can  shift  the  curve  in  the  direction  of  the 
F-axis,  Fig.  8,  and  the  values  of  f'(x)  will  remain  unchanged. 


52       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 

In  other  words,  the  curves  y  =f(x)  and  y  =f(x)  +  c,  where  c  is 
any  constant,  have  the  same  derived  curve   (see  Art.  24). 

EXERCISES 
Draw  the  graphs  of  the  following  functions  and  the  derived  curve  for  each 
of  them.     Study  carefully  the  combined  graphs,  and  trace  out  the  connection 
between  the  slope  of  the  original  curve  and  the  ordinate  of  the  derived  curve. 

1.   V  =  —  r-  X.  2.    y  =  x*  -  2  x2  ■+  5. 

9      12  J 

3.    y*  =  8x-10.  4-    ocy  =  25. 

5.   y  =  3  +  v^-4.  6.    p  =  8  +  (as  -  4)$. 

7.  Draw  several  curves,  and  by  observing  the  variation  in  the  slope  and 
the  turning  points  sketch  in  approximately  the  derived  curves. 

8.  Draw  a  curve  at  random,  and  by  observing  the  variation  of  the  ordi- 
nate draw  roughly  the  curve  of  which  the  first  curve  is  the  derived  curve. 

36.  Rolle's  theorem.  The  following  proposition,  known  as 
Rolle's  theorem,  is  essential  in  the  development  of  certain  other 
useful  theorems. 

Theorem.  If  f(x)  and  f'(x)  are  single-valued  and  continuous 
for  all  values  of  x  from  x  =  a,  to  x  =  b,  and  iff(a)  =f(b)  =  0,  then 
f'(x)  vanishes  for  at  least  one  value  of  x  between  a  and  b* 

Either  f(x)  has  a  constant  value  zero  for  all  values  of  x  between 
a  and  b,  or  it  varies  with  x.     In  the  first  case  f'(x)  is  zero  for  all 
values  of  x.     In  the  second  case,  since 
f(a)=f(b)  =  0, 

f{x)  must  at  some  point  begin  to  increase  and  afterwards  decrease, 
or  vice  versa.  It  must  then  have  a  turning  point  for  some  partic- 
ular value  ot  x,  say  x  =  xu  lying  between  a  and  b,  since,  by  hy- 
pothesis, f(x)  is  continuous. 

Geometrically,  Rolle's  theorem  means  that  if  a  continuous  curve 
cuts  the  X-axis  in  two  points  x  =  a,  x  =  b}  and  has  a  definite 
direction  at  every  point  in  this  interval    (a,  b),  then  at  some 


*  Rolle's  theorem  is  stated  here  in  sufficiently  general  terms  for  our 
present  purposes.  The  proof  given  holds  for  the  theorem  as  stated.  For  a 
more  general  statement  of  the  theorem  and  its  proof,  see  Pierpont's  Theory 
of  Functions,  Vol.  I,  p.  246. 


Arts.  36,  37] 


LAW   OF   THE   MEAN 


53 


intervening  point,  say 
x  =  xx,  the  tangent  to  the 
curve  is  parallel  to  the 
X-axis  (Fig.  13). 

It  is  at  once  evident 
that,  instead  of  the  con- 
dition f(a)=f(b)  =  0, 
we  might  have  f(a)  and 
f(b)  equal  to  any  con- 
stant so  long  as  they  are  equal  to  each  other, 
remains  precisely  the  same  in  the  two  cases. 


+-X 


The  argument 


EXERCISES 

1.  By  Rolle's  theorem  show  that  at  least  one  real  root  of  the  equation 
/'  (x)  =  0  lies  between  any  two  real  roots  of  the  equation  /  (x)  =  0. 

2.  From  Ex.  1  show  that  if  two  roots  of  f(x)  =  0  are  equal,  one  root  of 
/'  (x)  =  0  coincides  with  them.    Give  a  geometric  illustration  of  this  statement. 

37.  Law  of  the  mean.  By  the  use  of  Rolle's  theorem,  we  may 
deduce  one  of  the  most  fundamental  theorems  of  the  differential 
calculus,  known  as  the  law  of  the  mean,  or  the  theorem  of  mean  value. 

Theorem.  Let  f(x)  and  f'(x)  be  single-valued  and  continuous 
functions  of  x  in  the  interval  a<x<b.  Then  there  exists  at  least 
one  value  x0  of  x  for  which 


f(p)-f{o)_f(. 


a  <x0  <  b. 


Without  loss  of  generality,  we  may  assume  f(a)  <  /(&).     Geo- 


metrically, the  quotient 


/(&)  -/(«) 


gives  the  slope  to  the  secant 


AB,  Fig.  14 


b  —  a 

Let  y  =f(x)  be  represented  by  the  curve  APDB, 
B  passing  through  the  points   A 

and  B ;  f'(xo)  is  then  the  slope 
of  the  tangent  to  this  curve  at 
the  point  D  whose  abscissa  is 
se0.  The  theorem  asserts  geo- 
metrically that  there  exists  at 
least  one  point  D  of  the  curve 
at  which  the  tangent  is  parallel 
to  the  secant  line  AB. 


54       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 

To  prove  this  theorem  we  proceed  as  follows.  The  equation 
of  the  line  AB  is  of  the  form 

y  =  m(x  —  a)  +  c, 

where  the  slope  m  is  -'^  \  ~~JW  an(j  the  ordinate  c  of  the  point 
b  —  a 

where  the  curve  cuts  the  line  x  =  a  is  f(a).     Hence,  we  have 

y  =mbZfa(a)  (• ~ a)  +/(«)•  W 

The  distance  of  any  point  P  on  the  curve  from  the  line  AB,  when 
measuring  along  the  ordinate  through  P,  is  given  by 

if,(x)  =  PQ  =  MQ-MP 

=mbZfa(a)  (*  - a)  +/(a)  -/<*)•         (2) 

Now  i/^(a?)  is  a  function  that  satisfies  all  of  the  requirements  of 
Eolle's  theorem.  At  some  point,  say  x  =  x0,  of  the  interval  (a,  b), 
we  have  therefore 

f  (z0)  =/(*>)-/(«)  _/(a.o)  =  o,  (3) 

o  —  a 

whence  /(*%  ~{(^  =  /W,  (4) 

as  the  theorem  requires. 

The  theorem  may  be  stated  in  another  form  which  is  sometimes 
convenient.  The  fact  that  x0  lies  between  a  and  b  can  be  ex- 
pressed by  the  relation 

x0  =  a  +  0(b  —  a), 

where  0  is  some  number  between  0  and  1.     We  may  also  put 

b  —  a  =  h. 

Equation  (4)  then  takes  the  form 

f(a  +  h)-f(a)=f(a  +  Qh^  (5) 

ft/ 

The  fraction  ^  '  ~~J^a)  evidently  measures  the  average  rate  of 
b  —  a 

increase  of  the  function  in  the  interval  b  —  a.      Thus,  suppose 

«!  and  s2  denote  distances  traversed  by  a  moving  point  in  the 


Art.  38]  EQUATION   OF  THE   TANGENT  55 

times  tx  and  t2  respectively ;  the  fraction  zSHh  gives  the  average 

t2      £1 

speed  of  the  point  for  the  time  interval  t2  —  tx,  and  the  law  of  the 

mean  asserts  that  at  some  instant  within  this  interval  the  actual 

instantaneous  speed  of  the  point  is  equal  to  the  average  or  mean 

speed  for  the  whole  interval. 

EXERCISES 

1.  Show  geometrically  that  Kolle's  theorem  is  a  special  case  of  the  law 
of  the  mean. 

2.  If  f(x)  =  x2,  find  the  value  of  6  that  satisfies  relation  (5)  :  (a)  when 
a  =  6,  h  =  1 ;  (6)  when  a  =  12,  h  =  4. 

3.  If  f(x)  =  x3,  find  the  value  of  6  in  order  that  (5)  shall  be  satisfied  when 
a  =  4,  h  =  1. 

4.  A  smooth  curve  is  passed  through  three  points  (6,  3),  (8,  5),  and 
(10,  8.5),  which  have  been  determined  experimentally,  and  the  slope  of  the 
curve  at  the  intermediate  point  (8,  5)  is  desired.  Find  by  the  law  of  the 
mean  an  approximate  value  for  this  slope. 

38,   Equation  of  the  tangent  and  of  the  normal  to  a  curve.     Let 

(x1}  2/j)  be  any  point  on  the  curve  whose  equation  is 

y  =/(*)• 

As  we  have  seen,  the  slope  of  the  tangent  to  the  curve  at  the 
point  (xl9  ?/i)  is  given  by  substituting  the  value  as,  for  x  in  the 
derived  function  /'(a?);  that  is,  for  x  =  xx,  t&n  cf>=f'(x1).  From 
analytic  geometry,  the  equation  of  the  tangent  at  any  point 
(xd  Vi)  on  *ne  curve  is 

V-Vi  =  rn(x  -  xx), 

where  m  denotes  the  slope  of  the  tangent  at  the  point  in  question. 
Replacing  m  by  f'(xi)>  we  have  for  the  equation  of  the  tangent  to 
the  curve  in  terms  of  the  derivative 

y~y1=//(aJ1)(«-aJ1).  (1) 

The  normal  to  a  curve  is  perpendicular  to  the  tangent  to  the 
curve  at  the  point  of  tangency.  The  condition  that  one  straight 
line  shall  be  perpendicular  to  another  is  that  the  slope  of  the  one 
shall  be  the  negative  reciprocal  of  that  of  the  other.     It  follows 


56       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 

that  the  equation  of  the  normal  to  the  given  curve  at  the  point 
(»i,  2/i)  is 

*-**-- rbtf*'m*  (2) 

Because  of  the  relation 

*"ii'm'  (Art32) 

the  slope  of  the  normal  can  frequently  be  most  conveniently 
obtained  by  substituting  y  =  y1  in  Dvx. 

Ex.    Find   the   equations    of    the    tangent    and    normal    to    the    curve 
y  =  xi  at  the  point  (4,  8). 

We  have  /'(»)  =  !«*> 

and  for*  =  4,  /'(*)  =  !  x  2  =  3. 

Substituting  this  value  of  /'(as)  in  (1),  the  equation  of  the  tangent  is 

y-8  =  3(a:-4), 

or  3  x  —  y  =  4. 

At  the  given  point,   —  =  -.     Hence,  the  equation  of  the  normal  is 

f'(xi)      3 

or  x  +  3  y  =  28. 

EXERCISES 

Find  the  equations  of  the  tangent  and  normal  to  the  following  curves  at  the 
points  indicated. 

1.  y*  =  3x  +  10,  at  (-2,  4). 

2.  xy  =  36,  at  (4,  9). 

,  at  (a,  a). 


2a  —  x 

4.  (x  -  4)2  +  (y  +  3)2  =  100,  at  (10,  5). 

5.  —  +  -^=1,  at  (2,  4\/2l). 
25      16  V     3         J 

6-  y  =  x*  -  3  x  +  10,  at  (2,  12). 

7.  y  =  a.i_I,  at(2,8|). 

Find  the  equations  of  the  tangent  and  normal  to  the  following  curves  at  the 
point  (xi,  yi). 

8.  y8  =  ax*.  9.    05$  +  ?/*  =  a*. 

10.  xn  +  yn  =  an.  11.    ^--£-  =  1. 

a2     62 


Arts.  38,  39] 


LENGTH   OF   TANGENT,    ETC. 


57 


39.  Lengths  of  tangent,  normal ;  subtangent,  subnormal.  The 
length  of  the  part  of  the  tangent  to  a  curve  that  lies  between  the 
point  of  tangency  and  the  axis  of  abscissas  is  called  the  length  of 
the  tangent.  The  length  of  that  part  of  the  normal  intercepted 
between  the  same  point  on  the  curve  and  the  axis  of  abscissas  is 
called  the  length  of 
the  normal.  The 
projections  of  these 
lengths  upon  the 
axis  of  abscissas  are 
called  respectively 
the  subtangent  and 
the  subnormal.  Thus 
in  Fig.  15  let  the 
tangent  to  the  curve 
at  the  point  P,  hav- 
ing the  coordinates 
x1}  ylf  meet  the  X-axis 
in  the  point  A,  and  let  the  normal  at  the  point  P  meet  it  at 
C.  Let  BP  be  a  perpendicular  let  fall  from  P  upon  the  X-axis. 
Then  for  the  point  P,  PA  is  the  length  of  the  tangent,  PC  is  the 
length  of  the  normal,  and  AB  and  BC  are  the  subtangent  and 
subnormal  respectively. 

In  the  triangles  APB  and  BPC  the  side  BP  =  yx  is  given,  as  is 
also  the  angle  PAB  —  CPB  =  <f>.  If  the  equation  of  the  curve  is 
y  =  f(x),  tan  <fi=f'(x1).  The  lengths  of  the  tangent  and  normal, 
the  subtangent  and  the  subnormal,  may  therefore  be  expressed  in 
terms  of  y,  and/' (ajj).     For  example, 


since 
we  have 


AB  =  BP  cot  <f>, 
subtangent  =  yx  cot  <f>  =  Jf1  * 


Ex.   Find  the  length  of  the  normal  to  the  curve  y2  =  —  at  the  point  (4.  4). 

4  i 

From  Fig.  15,  the  required  length  is  yx  sec  <p.     Since  Dxy  =  |x7, 


we  have 
therefore 


/'(zi)=/'(4)=fV4:=f; 
tan  <f>  =  |,    sec  <f>  =  Vl  +  (f  )2, 


and  the  length  of  the  normal  is  4Vl  +  (|)2  =  2  Vl3. 


58       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES  [Chap.  III. 

EXERCISES 

1.  Find  the  length  of  the  tangent,  the  subtangent,  and  the  subnormal  for 

xH 
the  curve  y2  =  — . 
4 

2.  Derive  general  formulas  for  the  length  of  the  tangent,  length  of  the 
normal,  subtangent,  and  subnormal  in  terms  of  yx  and/'(xi). 

Find  the  subtangent,  subnormal,  length  of  the  tangent,  and  length  of  the 
normal  for  each  of  the  following  curves  at  the  points  indicated. 


3. 

y2  =8 *,  at  (2,  4). 

4. 

xy  =  a2,  at  (xtl  y{). 

5. 

(x  -  4)2  +  (y  +  3)2  =  25,  at  (7,  1) 

6. 

y   =  x*  -  3  x  +  10,  at  (xi,  yi). 

7. 

y%  —  ax^  at  («}  a). 

« 

r3 

2a  —  x 

9.   xff  +  y*  =  o«,  at  (xi,  jfi). 

10.  Show  that  the  subnormal  to  the  parabola  y2  =  2  mx  is. constant. 

11.  Show  that  the  length  of  the  normal  to  the  curve  x2  +  y2  =  a2  is 
constant. 

12.  Using  the  equations  of  the  tangent  and  the  normal  in  Art.  38, 
derive  formulas  for  the  length  of  the  tangent,  length  of  the  normal,  etc.,  by- 
finding  the  intercepts  of  these  lines  on  the  X-axis. 


40.   Tan  ty,  cot  ^.     Let  p,  0  be  the  polar  coordinates  of   any 
point  on  a  curve,  0  being  taken  as  the  independent  variable.     In 

order  to  determine  the  di- 
rection of  the  curve  at  any 
point,  it  is  convenient  to 
express  the  tangent  and 
cotangent  of  the  angle  if/ 
between  the  radius  vector 
and  the  tangent  to  the 
curve  in  terms  of  p,  6,  and. 
their  derivatives. 

In  Fig.  16,  let  P  be  any 

point  on  the  given  curve, 

If  0  takes  an  increment  A0, 

Let  P'  be  the 


Fig.  16. 


and  let  its  polar  coordinates  be  p,  0. 

then  p  will  have  a  corresponding  increment  Ap. 


Art.  40]  TAN  $,    COT  +  59 

point  having  the  coordinates  0  -f-  A0,  p  4-  Ap,  and  let  MP  be  drawn 
perpendicular  to  OP'. 

From  the  figure,  we  have 

MP  =  p  sin  A0,     OM  =  P  cos  AO, 

and  MP'  =  OP'  —  p  cos  A0  =  p  +  Ap  —  p  cos  A<9. 

Hence,  tan  3fP'P  = psinA^ 

p  +  Ap  —  p  cos  A0 

As  A0  =  0,  the  angle  MP'P  approaches  the  angle  if/.     Hence  we 
have 

L     tan  MP'P=     Ln P^M 

A*  =  0  A0  =  0  Ap  +  p(l  -  cos  A0)' 

sin  Afl 

or  tan  ^  =     JJ 

r      A0  =  0  Ap         /l  -  cos  A(9 

A0     P\       AO 

But  we  have  seen  that 

hence,  we  have 

tan*  =  -^  =  pDp0.  (1) 

From  (1)  it  follows  at  once  that 

col!^      l_  =  3l>p.  (2) 

T      taw  \|/      p      * 

Ex.     Let  the  equation  of  a  curve  be  p2  =  -  .     Find  the  angle  \p  for  p  =  1. 

Writing  the  given  equation  in  the  form 

0  =  ^  =  ap-2, 
P2 
we  find  Dp0  =  -  2  ap"3. 

Hence,  tan  $  =  p  DPe  =  -2  ap~2  =  -  2  - , 

P2 

and  cot  4>  =  —  ■£—  - 

Y  2a 

Forp  =  l,  ta,n\(/=  —  2a, 

whence  f  =  arc  tan  (-2  a). 


60      ELEMENTARY  APPLICATIONS  OF  DERIVATIVES    [Chap.  III. 

41.  Length  of  polar  tangent  and  polar  normal ;  polar  subtangent, 
polar  subnormal.     Given  a  curve  and  the  polar  coordinates  (pl7  02) 

of  any  point  P  upon  it  (Fig. 
17).  At  the  point  P,  draw 
the  tangent  and  the  normal 
to  the  curve  and  through  the 
pole  0  draw  a  perpendicular 
to  the  radius  vector  OP.  Let 
this  perpendicular  cut  the  tan- 
gent and  the  normal  in  the 
points  T  and  N  respectively. 
PN  is  called  the  length  of  the 
polar  normal,  and  PT  the 
length  of  the  polar  tangent.  The  projections  of  these  lines  upon 
the  perpendicular,  namely,  NO  and  OT,  are  called  the  polar  sub- 
normal, and  polar  subtangent,  respectively. 

It  will  be  observed  that  in  the  case  of  rectangular  coordinates, 
the  line  upon  which  the  tangent  and  normal  were  projected  to 
determine  the  subtangent  and  subnormal  was  the  X-axis,  while 
in  polar  coordinates,  it  is  not  the  initial  line  but  a  line  through 
the  pole  perpendicular  to  the  radius  vector. 

In  each  of  the  right  triangles  OPT  and  OPN  one  angle  if/  and 
one  side  OP=px  are  given.  Since,  from  the  preceding  article, 
ij/—  arc  tan^P'^),  where  P'(0i)  is  obtained  from  the  equation 
of  the  curve  p  =  F(6),  it  follows  that  the  four  lengths  in  question 
can  be  expressed  in  terms  of  px  and  P'(#i). 

a2 
Ex.     Find  the  lengths  of  the  polar  tangent  of  the  curve  p2  =  —  ■ 

Putting  the  equation  in  the  form  p  =  ad~2, 


we  get 

J)flp=-lae  *;   hence,  F'(dl)= 2L-. 

2dfi 

Therefore 

cotxl/           1      «             6\t     a               1 

and 

tarn//  =-  2  0i. 

From  Fig.  17,  length  of  polar  tangent  =  pi  sec  \f/  =  —  Vl  +  4  0i2. 


Arts.  41,  42]  SPEED  AND  ACCELERATION  61 


EXERCISES 

1.  Derive  general  formulas  for  the  lengths  of  the  polar  tangent  and  polar 
normal,  the  polar  subtangent,  the  polar  subnormal,  in  terms  of  pi  and  F'(d{). 

2.  Find  the  polar  subtangent  and  polar  subnormal  of  the  curve  p2  =  —  • 

8 

3.  Find  the  length  of  the  polar  normal  of  the  same  curve. 

Find  general  expressions  for  the  length  of  the  tangent,  subtangent,  length 
of  the  normal,  and  subnormal  of  the  following  curves  : 

4.  P  =  ad.  5.   p  =  ad2.  6.   p  =  1  +  -.  7.    p2  =  ad  +  bd2. 

d 

8.  Find  tan  \f/  for  the  curves  of  Exs.  4-7. 

9.  Show  that  the  polar  subtangent  to  the  curve  pd  =  a  is  of  constant  length. 
Trace  the  curve. 

42.  Speed  and  acceleration.  Let  a  point  move  in  a  given  path, 
straight  or  curved,  and  let  s  denote  the  variable  distance  of  the 
point  measured  along  the  path  from  some  fixed  origin  0.  Assum- 
ing continuous  motion  from  0,  the  distance  s  evidently  depends 
upon  the  time  t ;  that  is,  s  =  f(t).     In  a  time  interval  At  the  point 

As 
moves  over  an  element  of  path  of  length  As,  and  the  ratio  —  gives 

the  mean  speed  of  the  point  for  this  interval.  The  limiting  value 
of  this  quotient  as  the  time  interval  A£  approaches  zero  gives  the 
instantaneous  speed  at  the  beginning  of  the  interval.  Denoting 
this  by  v,  we  have  therefore 

v=     L  ^  =  Dts;  (1) 

that  is,  the  speed  of  a  moving  point  is  the  time-derivative  of  the  space 
traversed. 

The  words  "speed"  and  "velocity"  are  often  used  as  synonyms.  An 
important  distinction  is,  however,  frequently  made.  The  term  "speed"  is 
used  to  indicate  merely  the  rate  of  motion  in  the  path  irrespective  of  direc- 
tion. Speed,  therefore,  has  only  magnitude.  The  term  "velocity"  carries 
with  it  the  additional  notion  of  direction,  hence  to  specify  a  velocity  both 
magnitude  and  direction  are  required. 

The  speed  v  is,  in  general,  a  function  of  the  time  t.  If  Av 
denotes  the  change  of  speed  between  two  points,  then  the  quotient 


62      ELEMENTARY  APPLICATIONS  OF  DERIVATIVES    [Chap.  III. 

—   defines   the   average    tangential   acceleration*    between   those 
At 

points  ;  and  the  limit  of  this  quotient  as  At  =  0  defines  the  instan- 
taneous tangential  acceleration.     Denoting  this  by  a,  we  have 

a=    L    %-Dp\  (2) 

that  is,  the  tangential  acceleration  is  the  time-derivative  of  the  speed. 
The  ordinary  unit  of  speed  is  the  foot  per  second,  that  of  accel- 
eration the  foot  per  second  per  second.     These  are  abbreviated  to 
ft./sec.  and  ft./sec.2  respectively. 

Ex.   In  a  certain  motion  the  space  described  is  expressed  as  a  function  of 
the  time  by  the  following  equation  : 

s  =  at'2  +  bt  +  c. 
For  the  speed,  we  have  v  =  Dts  =  2  at  +  6, 

and  for  the  tangential  acceleration,  we  obtain 

a  —  B{0  —  2  a. 

43.   Angular  speed  and  acceleration.     If  a  body  rotates  about  a 
fixed  axis,  any  given  point  of  it,  not  on  the  axis,  moves  in  a  circle 

whose  center  lies  on  this  axis, 
and  whose  plane  is  perpendicu- 
lar to  the  axis.  Let  0  be  the 
center  and  PAB  the  circular 
path  of  the  point,  Fig.  18.  Dur- 
ing the  motion  of  the  point  from 
P  to  A,  the  radius  OA  sweeps 
over  an  angle  6,  and  in  the  ad- 
FlG-  18,  ditional  interval  of  time  At  re- 

quired for  the  motion  from  A  to  B  it  sweeps  over  the   angle 

AOB  =  A8.     The  ratio  —  defines  the  mean  angular  speed  of  the 
body  between  the  positions  OA  and  OB,  and  the  limit  of  this 


*  Tangential  acceleration  is  the  acceleration  in  the  direction  of  the  motion, 
that  is,  tangent  to  the  path.  In  the  case  of  rectilinear  motion,  this  is  the 
only  acceleration  ;  but  in  the  case  of  a  point  moving  in  a  curve,  there  is 
another  acceleration  perpendicular  to  the  tangent. 


Art.  43]  ANGULAR   SPEED   AND   ACCELERATION  63 

ratio  as  B  is  made  to  approach  A  defines  the  instantaneous 
angular  speed  for  the  position  OA  of  the  radius.  Denoting  this 
angular  speed  by  a>,  we  have  « 

o>  =    L    ~  =  Dt0;  (1) 

At  =  0  A* 

that  is,  the  angular  speed  is  the  time-derivative  of  the  angle  swept 
over. 

The  angular  speed  may  be  constant  or  variable.  If  variable, 
the  increment  between  two  positions,  say  A  and  B,  may  be  de- 
noted by  Aw,  and  the  ratio  — —  gives  the  mean  angular  acceleration 

between  the  positions  in  question.  The  limit  of  this  mean  value 
as  the  two  chosen  positions  are  made  to  approach  each  other  is 
the  instantaneous  angular  acceleration,  which  is  denoted  by  a.  We 
have  then 

a=     L     77-A-3  (2) 

At  =  0  At 

that  is,  the  angular  acceleration  is  the  time-derivative  of  the  angular 
speed. 

With  angles  measured  in  radians,  the  unit  of  angular  speed  is 
the  radian  per  second  (rad./sec),  and  that  of  angular  acceleration 
is  the  radian  per  second  per  second  (rad./sec.2). 

The  relation  between  the  angular  speed  of  a  rotating  body  and 
the  linear  speed  of  any  point  of  the  body  in  its  circular  path  is 
readily  derived.  Eeferring  to  Fig.  18,  AB  is  an  element  As  of 
the  circular  path  of  P.  Denoting  the  radius  OA  by  r,  we  have 
therefore 

As  =  rAO, 

whence  —  =  r — , 

At        AT 

and  finally  Dts  =  rDt6.  (3) 

That  is,  the  speed  of  any  point  of  a  body  rotating  about  a  fixed  axis 
is  the  product  of  the  distance  of  the  point  from  the  axis  and  the  angu- 
lar speed  of  the  body. 


64      ELEMENTARY  APPLICATIONS  OF  DERIVATIVES    [Chap.  III. 

A  similar  law  holds  for  tangential  acceleration.     Thus  from  (3) 

v  =  rut  j 
whence  Dtv  =  rDtio, 

or  a  =  ra ;  (4) 

that  is,  the  tangential  acceleration  of  the  'point  is  the  angular  accel- 
eration about  the  fixed  axis  multiplied  by  the  distance  from  the  point 
to  the  axis. 

EXERCISES 

1.  If  a  body  is  projected  vertically  upward  with  a  speed  of  v0  feet  per 
second,  the  space  traversed  in  t  seconds  from  the  instant  of  projection  is  given 
by  the  equation 

8  =  vtf  -  I  gt2. 

(a)  Find  expressions  for  the  speed  and  acceleration  at  the  time  t\. 
(b)  Take  v0  =  800  and  g  =  32.2,  and  find  the  speed  and  acceleration  at  the 
end  of  2  seconds ;  (c)  at  the  end  of  12  seconds. 

2.  In  Ex.  1,  find  the  whole  time  occupied  by  the  body  rising  and  falling, 
also  the  height  to  which  it  rises. 

Suggestion  :  Make  s  =  0  and  solve  for  t. 

3.  When  a  body  moves  in  a  straight  line  under  the  influence  of  an 
attractive  force  that  varies  as  the  inverse  square  of  the  distance,  the  motion 

is  given  by  the  equation  v2  =  -,  in  which  A:  is  a  constant.    Show  that  the 

acceleration  is — . 

2  s2 

4.  The  angle  (in  radians)  through  which  a  given  rotating  body  turns,  start- 
ing from  rest,  is  given  by  the  equation 

0  =  t2  +  5  t. 
Find  the  angular  speed  and  angular  acceleration  at  the  end  of  4  seconds. 

5.  If  the  angle  is  given  by  the  equation 

d=\\2t-  16*2, 
find  (a)  the  speed  and  acceleration  at  the  end  of  2.5  seconds ;  (6)  the  time 
that  elapses  before  the  body  comes  to  rest. 

Derive  expressions  for  w  and  a  from  each  of  the  following  relations  between 
B  and  t : 

6.   6  =  at-  bfi.  7.    e  =  aA  8.    6  =  a  +  bt  +  cfi. 

44.  Miscellaneous  applications.  In  physics  and  in  chemistry 
the   notion   of    the   derivative    is    repeatedly    encountered.      In 


Art.  44]  MISCELLANEOUS  APPLICATIONS  65 

the  case  of  a  moving  point,  the  term  "  speed "  is  used  for  the 
time-derivative  of  the  distance  traversed.  In  chemistry,  likewise, 
we  have  the  same  term  used  for  other  time-derivatives ;  thus,  the 
speed  of  reaction  and  the  speed  of  solution  are  such  derivatives. 
In  physics  we  meet  with  a  great  number  of  derivatives  expressing 
the  rates  of  change  of  various  physical  magnitudes.  The  heating 
of  substances,  variations  of  pressure,  density,  and  temperature,  the 
variations  in  velocity  and  in  energy,  etc.,  are  such  changes.  A 
few  of  these  derivatives  are  discussed  in  the  following  paragraphs. 

(a)  Coefficients  of  expansion.  Let  a  rod  or  wire  have  unit  length  (1  foot 
or  1  yard)  at  some  standard  temperature,  say  32°  F.  or  0°  C.  When  heated  the 
rod  expands  by  an  amount  x  depending  upon  the  temperature.  Thus,  denoting 
the  temperature  by  t  ,  the  expansion  is  x  —  /(r )  and  the  new  length  of  the  rod  is 

1  +  s  =  1+/(t).  (1) 

If  now  the  temperature  rises  by  an  amount  At,  the  length  of  the  rod  will 

increase   by  a  corresponding   amount  Ax.    The  quotient  —  is  called  the 

At 
average  coefficient  of  linear  expansion  for  the  interval  At,  and  its  limit  as  At 
is  made  to  approach  zero  is  the  coefficient  of  linear  expansion  for  the  tempera- 
ture t.     Denoting  this  by  (7i,  we  have 

d=    L    ^-=Drx.  (2) 

At  =  0At 

In  most  cases  we  may  assume  with  sufficient  approximation 

x  -/(t)  =  ar  +  br2, 
whence  (J\  =  a  +  2  br ;  (3) 

that  is,  C\  is  itself  a  function  of  the  temperature. 

In  the  same  way  we  may  arrive  at  the  coefficients  of  superficial  and  cubical 
expansion.  Consider  a  cube  with  its  edge  having  unit  length  at  0°  C. 
Assuming  equal  expansion  in  all  directions,  each  edge  at  the  temperature  r  will 
have  a  length  1  +/(t).     Hence  the  area  of  a  face  will  be 

^  =  [1+/(t)]2,  (4) 

and  the  volume  of  the  cube  will  be 

V={l+f{r)y.  (5) 

The  derivatives  DTA  and  DT  V  are  the  coefficients  of  superficial  and  cubical 
expansion  respectively,  and  may  be  denoted  by  C2  and  C3. 

(6)  Specific  heat.  A  body,  originally  at  some  definite  temperature  t0,  is 
heated  and  its  temperature  rises.  The  heat  Q  absorbed  by  the  body  is  in 
general  a  function  of  the  final  temperature  r  ;  that  is, 

Q=f(r).  (6) 


66      ELEMENTARY  APPLICATIONS  OF  DERIVATIVES    [Chap.  III. 

An  increment  of  heat  AQ  causes  a  corresponding  rise  in  temperature  At,  and 

the  quotient  — *  is  called  the  average  specific  heat  of  the  body  for  the  inter- 

At 

val  At.     The  limit  of  this  quotient  as  At  approaches  zero  is  defined  as  the 
specific  heat  at  the  temperature  r.     Denoting  this  by  c,  we  have 

c=    L    ^  =  DTQ.  (7) 

At=0  At 


EXERCISES 

1.  Regnault's  experiments  on  the  heating  of  various  substances  are  repre- 
sented by  the  following  equations  : 

Ether,  Q  =  0. 5290  t  +  0. 000296  t2. 

Chloroform,  Q  =  0.2:324  r  +  0.00005  t2. 

Bisulphide  of  carbon,    Q  =  0.2352  t  +  0.000082  t2. 
In  each  case  Q  denotes  the  heat  required  to  raise  the  temperature  of  the  sub- 
stance from  0°  C.  to  t°  C.     For  each  substance  find  the  specific  heat  at  20°  C. 

2.  It  is  found  by  experiment  that  the  volume  of  water  which  at  4°  C.has 
unit  volume  is  given  by  the  equation 

V  =  1  +  a  (t  -  4)2, 

where  r  denotes  the  temperature  of  the  water  and  a  =  0.00000838.    Find  the 
coefficient  of  cubical  expansion  when  r  =  0°  ;  also  for  r  =  20°. 

3.  The  electrical  resistance  B  of  a  wire  varies  with  the  temperature  r  of 
the  wire,  the  relation  being  expressed  by  B  =/(f).  (a)  What  is  expressed 
by  the  derivative  DTB  ?  (6)  Find  the  expression  for  this  derivative  from 
Callendar's  formula 

in  which  B0,  a,  and  /3  are  constants. 

4.  (a)  What  derivative  expresses  the  rate  of  the  rise  of  temperature  of 
a  gas  with  respect  to  the  pressure  ?  (6)  With  respect  to  the  volume  ? 
(c)  What  derivative  gives  the  rate  of  change  of  the  energy  IT  of  the  gas 
with  respect  to  the  temperature  r  ? 

5.  The  pressure  of  the  atmosphere  decreases  as  the  distance  from  the 
earth's  surface  increases,  and  the  rate  of  change  of  pressure  with  the  height 
is  proportional  to  the  pressure.     State  this  law  in  the  form  of  an  equation. 

MISCELLANEOUS   EXERCISES 

1.  The  equation  of  a  curve  is  4  a2y  =  x3  —  2  ax2  +  a3, 
(a)  Find  the  slope  f or  x  =  0  and  x  =  a. 

(6)  Find  the  points  where  the  curve  is  parallel  to  the  X-axis, 
(c)  Find  the  points  at  which  the  slope  of  the  curve  is  1. 


Art.  44]  MISCELLANEOUS   EXERCISES  67 

2.  Find  the  angle  at  which  the  circle  x2  -f  y2  =  8  x  intersects  the  curve 

9       2-x 

3.  Find  the  equation  of  the  tangent  to  the  hyperbola  xy  =  30  that  cuts 
the  X-axis  in  the  point  x  =  4.  Show  that  the  point  of  tangency  of  any 
tangent  to  this  curve  lies  midway  between  the  intersections  of  the  tan- 
gent with  the  X-  and  F-axes. 

Investigate  the  following  curves  by  means  of  their  derivatives  : 

4.  „  =  *(«_*)*.  s.  p  =  Tf^-  6.  ?  =  f^f  • 

7.  Draw  curves  showing  the  speed  v  of  a  body  falling  in  vacuo  (s  =  \  gt), 
(a)  with  values  of  t  as  abscissas,  (6)  with  values  of  s  as  abscissas.  Show 
that  the  subnormal  of  the  second  curve  is  numerically  equal  to  the  acceler- 
ation. 

8.  In  the  case  of  the  semicubical  parabola  ay2  =  xs  show  that 

27 

subnormal  =  (subtangent)2  ■ ■ 

8  a 

9.  (a)  Find  the  equation  of  a  curve  whose  polar  subnormal  is  constant. 
(6)  Find  a  curve  whose  polar  subtangent  is  constant. 

10.  Show  that  Rolle's  theorem  does  not  hold  for  f(x)  =  (x  —  1)*  —  1 
between  x  =  0  and  x  =  2.     Explain  why. 

11.  Apply  Eq.  (5),  Art.  37,  to  the  function  f(x)  =  Xs  -  6  x  -  8.  Find 
the  value  of  d  for  a  =  4,  h  =s  1  j  also  for  a  =  7,  h  =  2. 

12.  Show  that  for  every  quadratic  function  f(x)  =  ax2  +  bx  +  c,  Eq.  (5), 
Art.  37,  is  satisfied  when  0  =  \.  From  this  fact  deduce  a  geometric  property 
of  the  parabola. 

13.  If  Q  denotes  quantity  of  heat,  r  temperature,  s  length  or  distance, 
and  t  time,  express  in  words  what  is  meant  by  the  equation 

DtQ=-k-DsT, 
which  applies  to  the  flow  of  heat  along  a  bar. 

14.  Show  that  the  coefficients  of  superficial  and  cubical  expansion  are  re- 
spectively two  and  three  times  the  coefficient  of  linear  expansion,  very  nearly. 

15.  Let  m  denote  the  mass  of  a  body  moving  in  a  straight  line,  v  the 
speed,  and  a  the  acceleration  of  the  body,  F  the  constant  force  acting  on  it, 
and  s  the  distance  traversed.  From  mechanics  we  have  the  following  defi- 
nitions and  symbols : 

mv  =  momentum  of  body,  (M), 

ma  =  force  acting,  (  F) , 

Fs  =  work  of  force  F,  (  W). 
Show  that  F  =  Dt(mv)  =  DtM, 

and  also  that  F  =  DSW. 


68       ELEMENTARY  APPLICATIONS  OF  DERIVATIVES    [Chap.  III. 

16.  The  kinetic  energy  of  the  body  (Ex.  15)  is  given  by  the  expression 

T=\  mv2. 
(a)  Show  that  the  momentum  is  the  v-derivative  of  the  kinetic  energy.     (&) 
Show  that  the  time-derivative  of  the  kinetic  energy  is  the  product  Fv. 

17.  If  for  any  motion  a  curve  is  drawn  with  the  speed  v  as  ordinates  and 
the  distances  s  as  abscissas,  show  that  the  tangential  acceleration  for  any 
value  of  s  is  given  by  the  subnormal  to  the  curve  at  that  point. 

18.  The  equation  of  van  der  Waals, 


(*+$)  («-»)=<* 


gives  the  isothermal  curves  of  carbon  dioxide.  By  means  of  the  derivative  in- 
vestigate the  general  form  of  the  isothermal  curves  obtained  by  giving  C  dif- 
ferent constant  values. 

19.  The  following  values  of  corresponding  pressures  and  temperatures  of 
saturated  steam  are  taken  from  a  standard  table  : 

p    lb./sq.  in.  84  85  86  87  88 

t    temperature  F.     315.19°     316.02°    316.84°    317.65°     318.45° 
Making  use  of  the  law  of  the  mean,  find  approximately  the  value  of  the  de- 
rivative Dpt  forp  =  85  ;  also  for  p  =  87.     Take  6  =  0.5. 

20.  Rankine's  formula  for  long  columns  has  the  form 

a 

y 


1  +  6x2 

By  means  of  the  derivative  investigate  the  general  character  of  this  function, 
and  sketch  the  curve  that  represents  it. 


CHAPTER   IV 
THE  DIFFERENTIAL  NOTATION 

45.  The  derivative  as  a  rate.  The  fundamental  problem  of 
differential  calculus  is  the  measurement  of  the  rate  of  change  of 
the  function  with  respect  to  the  variable.  We  are  not  concerned 
so  much  with  the  actual  change  of  the  function  as  with  its  change 
per  unit  increase  of  the  variable,  or,  in  other  words,  its  rate  of 
change.  The  great  importance  of  the  derivative  lies  in  the  fact 
that  it  gives  a  precise  measure  of  this  rate  of  change.  Thus  the 
derivative  Dts  measures  the  rate  of  change  of  the  distance  s  with 
respect  to  the  time,  or,  more  briefly  stated,  the  time-rate  of  s ;  the 
derivative  Dt0,  the  angular  speed,  is  likewise  the  time-rate  of  the 
angle  0.  Again,  consider  the  case  of  heating  a  metal  bar.  The 
length  x  of  the  bar  is  a  function  of  its  temperature  t,  say  x  =/(t), 
and  the  derivative  Drx  gives  the  rate  of  change  of  length  with 
respect  to  the  temperature. 

That  the  derivative  measures  the  rate  of  change  of  the  function 
is  easily  seen.  If  Ay  denotes  the  change  of  the  function  corre- 
sponding to  a  change  Ax  of  the  variable,  the  quotient 

Ay  =  f(x  +  Ax)-f(x) 
Ax  Ax 

gives  the  average  rate  of  change  for  the  interval  Ax.  The  rate  at 
which  y  is  changing  with  respect  to  x  at  any  particular  value  of  x, 
as  x0,  is  the  limiting  value  of  this  quotient  as  Ax  approaches  zero, 
that  is, 

It  is  not  necessary  that  y  be  expressed  directly  in  terms  of  x  in 
order  that  we  may  discuss  their  comparative  rates  of  change.  In 
fact,  it  is  often  convenient  to  compare  the  rates  of  change  of  two 
functions  by  means  of  a  third  variable.     For  example,  we  may 

69 

i 


70 


THE   DIFFERENTIAL   NOTATION 


[Chap.  IV. 


compare  the  relative  rate  of  change  in  the  market  value  of  wheat 
in  Chicago  and  steel  rails  in  New  York  by  the  use  of  a  third  vari- 
able representing  money.  The  speed  of  a  locomotive  may  be 
compared  with  that  of  a  street  car,  or  the  relative  speed  of  two 
chemical  reactions  may  be  found  by  means  of  a  third  variable  rep- 
resenting time. 

The  conception  of  the  derivative  as  a  rate  is  fundamental  in  the 
developments  of  the  present  chapter. 

46.   Differentials.     In  the  preceding  chapters  we  have  employed 
the  notation  of  derivatives  explained  in  Art.  16.     We  shall  now 

introduce  another  nota- 
tion, that  of  differentials, 
which  for  certain  pur- 
poses is  more  convenient 
than  that  of  derivatives. 

Let  us  consider  first  the 
case  in  which  y  is  ex- 
pressed directly  in  terms 
of  x,  and  let  y  =f(x)  be 
a  continuous  function  rep- 
resented by  the  curve  in 
Fig.  19.  At  any  point  P 
of  the  curve  let  a  tangent 
be  drawn,  and  let  PS  rep- 
resent an  assumed  increment  Ax  of  the  independent  variable  x. 
The  corresponding  increment  Ay  of  the  function  is  represented 
by  SQ,  while  the  intercept  SM  between  PS  and  the  tangent 
represents  the  product  Ax  •  tan  <f>  =  f\x)  Ax.  Denoting  this 
product  by  dy,  we  have  therefore 

dy=f(x)Ax.  (1) 

In  order  to  make  the  notation  symmetrical,  it  is  customary  to 
denote  the  increment  of  a;  by  dx;  that  is,  to  write  dx  for  Ax. 
With  this  convention  (1)  becomes 

dy=f'(x)dx.  (2) 

In  equation  (2)  dx  is  called  the  differential  of  x,  dy  the  differential 
of  y,  and  f\x)  dx  the  differential  of  f(x).     We  have  therefore  the 


Fig.  19. 


Art.  46]  DIFFERENTIALS  71 

following  definitions :  The  differential  of  the  variable  x  is  merely 
an  assumed  increment  of  x  ;  and  the  differential  of  the  function  y 
is  the  product  of  the  derivative  f'(x)  of  the  function  and  the  dif- 
ferential dx  of  the  variable. 

Frequently  f(x)  is  called  the  differential  coefficient,  that  is, 
the  coefficient  of  the   differential   of   the   independent  variable. 

Having  assigned  a  value  to  dx  (=  Ax),  the  values  of  dy  and  Ay  are 
determined  by  the  nature  of  the  functional  relation  y=f(x). 
Usually  these  two  symbols  do  not  represent  the  same  value.  The 
value  of  dy  is  determined  by  f'(x)  dx,  that  is,  by  dx  •  tan  cf>,  and  is 
represented  in  the  figure  by  SB.  On  the  other  hand,  Ay  is  the 
change  in  the  function  f(x)  corresponding  to  the  change  Ax  of 
the  independent  variable,  that  is,  SQ  in  the  figure.  The  two 
symbols  dy  and  Ay  represent  the  same  value  when,  and  only  when, 
y  =f(x)  is  represented  by  a  straight  line. 

It  should  be  noted  that  dy  is  the  increment  that  the  ordinate 
would  have  if  the  rate  of  change  of  y  at  the  point  P  were  main- 
tained throughout  the  interval  Ax.  It  is  evident  also  that  if  Ax  is 
sufficiently  small  dy  is  an  approximation  to  the  increment  Ay,  and 
that  the  smaller  Ax  is  chosen,  the  closer  is  the  approximation. 

The  relation  (2)  between  dy  and  dx  evidently  holds  when  x  and 
y  are  given  as  functions  of  a  third  variable  t.  The  parametric 
representation  has  the  advantage  that  it  enables  us  to  show 
clearly  the  relation  between  differentials  and  rates  as  measured 
in  terms  of  a  common  variable.     Suppose  we  have  given 

«=*(*),  2/  =  <K0-  (3) 

Eliminating  t  in  the  second  equation  by  means  of  the  first,  we 

have  y  =/(#),  where  x  =  <f>(t).  (4) 

Hence  by  Art.  31,  we  have 

Dty=f'(x)Dtx.  (5) 

As  we  have  seen,  the  derivatives  Dtx,  Dty  express  the  rates  of 
change  of  the  variables  x  and  y  with  respect  to  t,  the  common 
variable  in  terms  of  which  both  x  and  y  are  expressed.  It  will 
be  observed  that  Dtx,  Dty  enter  homogeneously  into  equation  (5) 


72  THE   DIFFERENTIAL   NOTATION  [Chap.  IV. 

as  do  dx,  dy  in  equation  (2).     Moreover,  by  a  comparison  of  equa- 
tions (2)  and  (5),  we  have 

J  W      cto     Dtx  J 

We  may  then  say  that  the  rates  Dtx,  Dty  are  either  equal  to  the 
differentials  dx,  dy  or  proportional  to  them.  This  relation  may  be, 
and  often  is,  made  the  basis  of  the  definition  of  differentials  ;  and 
it  enables  us  to  write  the  derivatives  Dtx,  Dty  in  terms  of  the 
differentials,  or  vice  versa,  whenever  it  suits  our  purpose  to  do  so. 
Since  we  have  ^ 

it  follows  that  a  derivative  is  equal  to  the  quotient  of  two  differen- 
tials.    It  is  frequently  convenient  to  write  the  derivative  as  such 

a  quotient,  and  in  the  future  we  shall  employ  either  Dxy  or  -2  as 

dx 

best  suits  the  problem  under  discussion.     We  may  read  -*-  either 

ax 

"  the  derivative  of  y  with  respect  to  »,"  or  "  differential  y  divided 

by  differential  xP 

The  distinction  between  the  symbols  Dxy  and  -^  should,  how- 

ax 

ever,  be  carefully  noted.     While  both  lead  to  the  same  numerical 

result  and  may  therefore  be  used  interchangeably,  Dxy  indicates 

that  a  certain  operation  has  been  performed  upon  the  function  ?/ 

with  respect  to  the  independent  variable  x.     It  is  not  possible  to 

separate  this  symbol  into  two  parts.     On  the  other  hand,  -*  is 

ax 

merely  a  quotient  which  can  be  dealt  with  by  the  ordinary  rules 

of  algebra. 

47.  Kinematic  interpretation  of  differentials.  An  instructive 
illustration  of  the  use  of  differentials  is  afforded  by  the  velocity 
components  of  a  moving  point. 

Suppose  a  point  P  to  move  along  a  curve  m,  Fig.  20,  and  let  Px 
and  Py  be  the  projections  of  P  on  the  X-axis  and  F-axis  respec- 
tively. As  P  moves  on  the  curve,  the  projections  Px  and  Py  move 
on  the  axes.     The  velocity  of  Px  along  OX  is  evidently  the  time- 


Art.  47]   KINEMATIC  INTERPRETATION  OF  DIFFERENTIALS      73 


rate  of  change  of  the  abscissa  x,  and  is  therefore  given  by  the 
derivative  Dtx.  Likewise,  the  velocity  of  Py  along  the  F-axis  is 
Dty.  The  velocity  of  P  along  the  curve  has  the  direction  of  the 
tangent  PT,  and  its  magnitude  is 
given  by  the  derivative  Dcs.  Suppose 
that  the  velocity  of  P  is  such  that  in 
a  unit  of  time  P  would  be  carried 
along  the  tangent  from  P  to  T.  This 
displacement  PT  can  be  effected  in  the 
following  manner :  Move  P  along  a 
horizontal  line  from  P  to  A,  and  at 
the  same  time  move  the  horizontal  line  PA  vertically  to  BT. 
Evidently,  if  the  horizontal  and  vertical  motions  are  made  with 
constant  velocity,  the  point  P  will  move  along  PT  with  constant 
velocity.  Since  the  motion  takes  place  in  a  unit  of  time,  the 
displacements  PT,  PA,  and  PB  respresent  respectively  the  ve- 
locity in  the  path,  the  velocity  of  the  horizontal  motion  of  Px, 
and  the  velocity  of  the  vertical  motion  of  Py\  that  is,  PT=Dts, 
PA  =  Dtx,  and  PB  =  Dty. 

The  two  velocities  Dtx  and  Dty,  which  together  may  replace  the 
velocity  Dts  in  the  curve,  are  called  the  component  velocities  in  the 
direction  of  the  axes.  From  the  figure  the  following  relations 
between  the  velocity  in  the  curve  and  the  components  along  the 
axes  are  evident : 

PT2  =  PA2+  PB2, 
hence  (Dtsy  =  (Dtxf  +  (Dty)\  (1) 


Substituting  the  differentials  ds,  dx,  and  dy  for  the  time-rates,  we 
have 


Furthermore,  since 

and 

we  have 


ds    =  doc   +  dy 

PA  =  PT  cos  <£, 
PB  =  PTsm<f>, 

dx  =  ds  cos  <f> 
dy  =  ds  sin  <£ 


(2) 


(3) 


Ex.  1.    A  point  moves  in  the  parabola  y'2  =  8  x  with  a  constant  velocity  of 
5  units.     Find  the  components  along  the  axes  when  the  point  is  at  (12.5,  10). 


74  THE   DIFFERENTIAL  NOTATION  [Chap.  IV. 

From  the  given  equation,  we  get 

2  yDt  y  =  8  Dtx, 

or  2  y  dy  =  8  <£r, 

whence  dy=-  dx. 


But  ds  =  Vdx2  +  dy2  =  5, 

whence  dx\—  +1  =  5. 

*  y'2 

We  have  then  <fo  = ^ — 


Vl6  +  ^' 

Kn  or 

which  becomes  for  y  —  10,  dfx  =  =  —  ^/29  • 

Vll6      29 

and  dw  =  —  •  —  V29  =  —  V29 

10      29  v^y      29 vzy- 

If  a  point  moves  along  a  curve  whose  equation  is  given  in  polar 
coordinates,  it  is  convenient  to  resolve  the  velocity  along  the  curve 

into  components  along  and  perpendicu- 
lar to  the  radius  vector,  respectively. 
Thus,  in  Fig.  21,  the  velocity  repre- 
sented by  PT  may  be  resolved  into  com- 
ponents represented  by  PA  and  PB. 
p      ~~  As    before,    PT  =  Dts.      The    velocity 

component  PA  is  evidently  the  time-rate 
of  increase  of  the  radius  vector  OP=  p,  that  is,  PA  =  Dtp.  The 
component  PB  is  the  velocity  that  P  would  have  if  it  remained  at 
rest  on  the  radius  while  the  latter  rotated  about  the  pole  0; 
hence,  if  o>  denotes  the  angular  speed  of  OP  about  the  pole  0, 

PB=OP<o  =  p Dft.  (See  Art.  43) 

Since  PT2  =  PA2  +  PB2, 

(Dtsy  =  (DtPy+(pDt6y,  (4) 

or,  substituting  differentials  for  the  time-derivatives, 

ds2  =  dp2  +  p3  cWa.  (5) 


Art.  48]        DIFFERENTIATION   WITH  DIFFERENTIALS  75 

It  follows  also  from  the  figure  that 

_PdO 


tan  \p  =  — 


(6) 


dp  =  ds  cos  if/ 
p  dd  =  ds  sin  if/. 

Ex.  2.    A  point  moves  in  the  curve  p  =  ad  with  a  constant  velocity  m. 
Find  the  velocity  components  when  d  =  it. 

Since  p  =  ad, 

we  have  Dtp  =  aDt6, 

or  dp  — a  dd. 

Then  dp2  =  a2  dd2, 

ds2  =  dp2  +  p2dd2=  (a2  +  p2)  dd2, 

ds  =  m  =  (a2  +  p2)%  dd, 


dd  = 


Va2  +  p2 

whence  pdd=      mp      , 

Va2  +  p2 

and  cfc  =      ma 


\/«2  + 
For  d  =  t,  p  =  air-     Substituting,  we  obtain 


poo  —  t  velocity  perpendicular  to  OP, 

VI  +  IT2 


dp  =  — m       ,  velocity  along  OP. 


Vl  + 

48.  Differentiation  with  differentials.  The  passage  from  the 
derivative  to  the  differential  notation,  or  vice  versa,  is  effected 
by  the  defining  equation 

dy=f(x)dx.  (1) 

Thus,  if  y  =  ax2  +  b, 

f(x)  =  %ax, 
whence  dy  =  2  ax  dx. 


76  THE   DIFFERENTIAL  NOTATION  [Chap.  IV. 

Conversely,  if  we  have  given 

we  obtain  by  division, 

^L=f(x)  =  3x2  +  4. 
dx 

In  the  operation  of  differentiation,  we  may  therefore  employ 
either  of  the  following  methods :  (1)  we  may  obtain  the  deriv- 
ative directly  by  the  theorems  already  established;  or,  (2)  we 
may  derive  an  equation  in  the  form  (1)  which  is  homogeneous  in 
the  differentials  of  the  variables,  and  then  obtain  the  derivative 
by  division. 

The  general  theorems  of  differentiation  may  be  easily  stated 
in  terms  of  the  differential  notation.     All  that  is  necessary  is  to 

replace  Dxy,  Dxu,  Dxv,  etc.,  by  their  equivalents  -^,  — ,  — ,  etc., 

and  multiply  by  dx.     The  following  are  the  results  thus  obtained, 
which  the  student  may  easily  verify. 

(a)  Given  y  =  c,  dy  =  0. 

(b)  Given  y  =  x,  dy  =  dx, 

(c)  Given  y  =  (u  +  v  +  w),  dy=du  + dv +  dw. 

(d)  Given  yz=u  +  c,  dy  —  du. 

(e)  Given  y  =  uv,  dy  =  udv  +  v  du. 

(/)  Given  y  =  cu,  [  dy  —  c  du. 

,  \  r\-               u                       7        0 du  —  u  dv 
(g)  Given  ?/  =  -,  dy  = 

v  v 

(h)  Given  2/  =  ^,  dy  =  =£$». 

v  v2 

(i)  Given  y  =  u,  dy  =  nun  du. 

The  following  examples  will  illustrate  the  use  of  the  preceding  formulas  : 
Ex.  1.    y 


02 

Vl- 

dy 

-X2. 

=  Vl- 

x2  d  (ax)  -f  ax 

•  dVl-x* 
dx 

=  aVl 
_o(l- 

x2dx        ax 

VT- 
-  2  x?)  dx 

-x2 

vr 


Akt.  49]     DIFFERENTIATION   OF   IMPLICIT  FUNCTIONS  77 

Ex.  2.     y 


a2dx 


vx2- 

a2 

dy 

V&- 

-  a2  dx  — 

■x-dy/'x2- 

-a2 

(Vx2 

-a2)2 
x2dx 

Vx2- 

-  a2  dx  — 

v'x'2  —  a2 

V(x2-a2)3 


EXERCISES 

Differentiate  the  following  functions  and  express  the  results  in  differential 
form. 


1. 

x  —  5 

y  =  x7  (x  —  a)"5. 
j/  =  (a  +  bxn)m. 

2. 

4. 
6. 
8. 

10. 
12. 

x  Va2  -  x2 

3. 

5. 
7. 

Va2  +  x2 

*  =  =" 

i/  =  (1  —  x)  VI  +  x. 
x  —  a 

a'2  V2  ax  —  x'2 

(2  ax  -  x2)^  ( 
ax3 

9. 

(3  x4  -  2  x2  +  2)  V2  x2  +  1. 

Vx2  +  a2 

.1. 

(x2-3)v/x2  +  l. 

a'2x 

49.  Differentiation  of  implicit  functions.  It  frequently  happens 
that  the  relation  between  two  variables,  as  x  and  y,  is  conveniently- 
expressed  by  means  of  an  implicit  function.  In  such  cases  the 
derivative  f'(x)   can   be   found   by  differentiating   according   to 

the  general  formulas  just  given   and  finding  the  quotient  -^. 

ctx 

Another  method  will  be  discussed  in  Chapter  XIII.  The  expression 

for  the  derivative  will  generally  contain  both  variables,  and  its 

numerical  value  can  be  found  for  known  simultaneous  values  of  x 

and  y. 

Ex.  1.     Given  x3y  -  4xy2  +  6  x2  -  Sy  =  0, 

we  obtain  by  differentiation  (using  the  formula  for  the  product) 

(3x2ydx  -\-xsdy)  -  (4 y2 dx  +  8 xy dy)  +  12xdx-3a>  =  0, 

whence  dy  =_Sx2y  -  4y2 +  12x. 

dx  x3  —  8  xy  —  3 


78  THE   DIFFERENTIAL  NOTATION  [Chap.  IV. 


The  values  x  =  1,  y  =  1  satisfy  the  original  equation,  and  for  these  values 
the  derivative  has  the  value 

§£}       -11- 
dx. 


■1     =11  =  1-1; 
J  i,i     io 

that  is,  the  slope  of  the  curve  is  1.1  at  the  point  (1,  1). 


Ex.  2.  The  equation  representing  the  adiabatic  expansion  of  air  {i.e.  ex- 
pansion without  gain  or  loss  of  heat)  is  pvk  =  C,  where  k  and  C  are  constants. 
p  denotes  the  pressure  and  v  the  volume  of  the  air.  Find  the  rate  of  change 
of  pressure  relative  to  the  volume. 

Differentiating,  we  obtain 

vk  dp  +  kp&~x  dv  =  0, 

or  vdp  +  kp  dv  =  0, 

whence  -j-=—k-> 

dv  v 

EXERCISES 

dv 
Find  the  derivative  ~  for  each  of  the  following  functions : 

1.   xs  -  xy2  -  c3  =  0.  2.    b2x2  -  a2y2  =  a2b2. 

3.  x2  -f-  y%  =  c.  4.   x4  —  x*y  +  x2y2  +  2  xy*  =  0. 

5.  Find  the  general  expression  for  the  slope  of  the  tangent  to  the  conic 
section  whose  equation  is 

ax2  +  2  hxy  +  by2  +  2  gx  +  2  fy  -f  c  =  0. 

6.  Find  the  slope  of  the  circle  x2  +  y2  =  100  at  the  point  (  —  8,  6). 

7.  Find  the  slope  of  the  ellipse  9  x2  +  14  y2  =  50  at  the  point  (2,  1)  ;  at 
the  point  (  —  2,  1) . 

8.  Find  the  slope  of  the  curve 

x*y  -  5  x2y2  +  12y*  =  0 
at  the  point  (2,  1). 

9.  Derive  expressions  for  the  polar  sub  tangent  and  polar  subnormal  of 
the  curves : 

(a)P2-pd^=C.  (b)  P-p2d+ C  =  0. 

10.  Derive  general  expressions  for  the  derivative  =5  when  the  expansion 
of  a  gas  follows  the  following  laws,  respectively  :  v 

(a)  pmvn  s  C,  (w,  w,  and  C  constants)  ;  . 


Art.  50]  APPLICATIONS   OF  DIFFERENTIALS  79 

Find  5*  in  the  following  : 

11.    (1  +  x)2y  =  (1  -  x)y*  -  «».  12.   a^a  =  62(6_.a;). 

13.   ccVy  —  yy/x  =  C. 

50.   Applications  of  differentials.     Suppose  we  have  given 

y  =/(*), 

and  by  differentiation  we  obtain 

dy=f(x)dx.  (1) 

The  differentials  cfy  and  d#  may  be  taken  as  representing  the  deriva- 
tives Dty  and  Dtx,  where  t  denotes  any  third  variable.  If  we  con- 
sider the  variable  t  as  denoting  time,  then  dy  and  dx  represent  the 
time  rates  of  the  function  y  and  of  the  variable  x  respectively. 
Hence,  if  the  time  rate  of  x  is  given,  the  time  rate  of  y  is  found 
by  differentiation. 

Ex.  1.  Boyle's  law  for  the  expansion  of  air  is  expressed  by  the  equation 
pv  —  C.  At  a  given  instant  the  pressure  is  40  lb./sq.  in.,  the  volume  of  the 
air  is  8  cu.  ft.,  and  the  volume  is  increasing  at  the  rate  of  0.5  cu.  ft./sec.  At 
what  rate  is  the  pressure  changing? 

We  have  pv  =  C, 

whence  p  dv  +  v  dp  =  0, 

or  dp  =  —  ®  dv. 

v 
Since  dv  =  Dtv  =  0.5, 

dp=-  —  x  0.5  =-2.5. 

8 

Hence,  the  pressure  is  decreasing  at  the  rate  of  2.5  lb./sq.  in.  per  second. 

If  in  equation  (1)  the  differentials  are  replaced  by  the  incre- 
ments Ay  and  Ax,  the  result  is  an  approximate  relation 

Ay=f'(x)Ax,  (2) 

which  is  frequently  useful  in  finding  the  error  in  the  result  of  a 
computation  due  to  a  small  error  in  the  observed  data  upon  which 

the  computation  is  based.  The  relative  error  — ^  is  given  approxi- 
mately by  the  equation  * 

*L=fM±x.  (3) 

y     /(*) 


80  THE   DIFFERENTIAL   NOTATION  [Chap.  IV. 

Ex.  2.  The  area  A  of  a  circle  being  determined  from  a  measurement  of 
the  diameter  D,  find  the  relative  error  in  the  calculated  area  due  to  an  error 
in  the  measurement. 


Since 

A  =  \  ttZ)2, 

we  have 

dA  =  a  irD  ■  dD, 

whence  approximately 

AA  =iwD-  AD, 

and 

AA=lIDAD  =  2Ad 
A        lirD2                  d 

Hence  an  error  of  one  per  cent  in  the  measurement  of  the  diameter  gives 
approximately  an  error  of  two  per  cent  in  the  calculated  area. 

EXERCISES 

1.  A  point  moves  in  the  straight  line  5x  —  3  y  =  30  in  such  a  way  that 
the  r-component  of  its  velocity  is  6.  Find  the  X-component  and  the 
velocity  in  the  path. 

2.  Find  the  X-  and  F-components  of  the  velocity  of  a  point  that  moves 
in  the  line  3  x  +  4  y  =  12  with  a  speed  of  15  units. 

3.  Suppose  that  a  straight  wire  rotates  about  one  end  with  an  angular 
speed  of  3  rad./sec.  and  that  a  bead  on  this  wire  moves  along  it  with  a  speed 
of  8  ft./sec.  When  the  bead  is  2  feet  from  the  center  of  rotation,  (a)  what 
is  the  component  of  its  velocity  perpendicular  to  the  wire  ?  (b)  what  is  its 
velocity  in  its  path  ? 

4.  Find  the  polar  equation  of  the  curve  described  by  the  bead,  Ex.  3, 
(a)  when  the  angular  speed  w  of  the  wire  is  a  times  the  linear  speed  of  the 
bead  along  the  wire  ;  (6)  when  «  is  constant  and  the  speed  of  the  bead  on 
the  wire  varies  inversely  as  the  distance  from  the  center  of  rotation. 

5.  If  a  soap  bubble's  diameter  is  increasing  at  the  rate  of  \  in./sec.  when 
it  is  4  inches,  at  what  rate  is  the  inclosed  volume  increasing  ? 

6.  If  a  man  6  feet  in  height  is  walking  at  the  rate  of  3  mi./hr.  away 
from  a  lamp-post  30  feet  high,  at  what  rate  is  the  end  of  his  shadow  receding 
from  the  lamp-post  ? 

7.  The  time  of  a  complete  oscillation  of  a  pendulum  of  length  L  is  given 
by  the  formula  rr 

*  =  2ttV-. 

yg 

Find  the  rate  of  change  of  the  time  compared  with  that  of  the  pendulum's 
length. 

8.  Find  the  error  in  the  calculated  time  if  the  error  in  the  measurement 
of  the  length  of  the  pendulum  is  0.5  per  cent. 

9.  Find  an  expression  for  the  relative  error  in  the  volume  of  a  sphere 
calculated  from  a  measurement  of  the  diameter  if  there  is  an  error  in  the 
measurement. 


Art.  50]  MISCELLANEOUS  EXERCISES  81 

10.  From  the  formula  for  kinetic  energy  T  =  \  ra»2,  show  that  a  small 
change  in  v  involves  approximately  twice  as  great  a  relative  change  in  T. 

11.  An  engine  cylinder  has  a  diameter  of  12  inches.  At  what  speed  is 
the  piston  moving  when  steam  is  entering  the  cylinder  at  the  rate  of  18  cu. 
ft./  sec.  ? 

MISCELLANEOUS   EXERCISES 

Differentiate  the  following  functions,  using  differentials. 


1     2  x2  +  1  ^ ^  2      Vl+g'+Vl 

3  x»  x 


3.   5aji(a^-4)$(x  +  7)i.  4. 


Va  —  Va  —  x 

for  the  functions  defined  by  the  fo] 
dx 


Find  -^  for  the  functions  defined  by  the  following  equations. 


5.   xs  -  5  ax2y  +  7  ys  =  0.  6.   x2  -  y2  = 


Vl  -  y2 
7.    -**-=<,  8.    1 =  *. 

»  +  y  Vx2  +  y2 

9.  Find  expressions  for  the  velocity  components  of  a  point  moving  in  the 
parabola  y2  =  20  x  with  a  speed  of  12  in.  /sec.  Find  the  values  for  the  point 
(5,  10). 

10.  A  point  moves  in  the  straight  line  6  x  —  8  y  —  12  with  a  velocity  of 
5  ft. /sec.  Find  the  components  of  the  velocity  along  the  X-  and  V-axes, 
respectively. 

11.  A  point  moves  in  the  circle  x2  +  y2  =  36  with  a  velocity  of  8  ft.  /sec. 
Find  the  X-  and  F-components  when  the  point  is  at  (5,  Vll). 

12.  A  point  moves  along  the  curve  p  =  — -.     When  p  =  3,  the  component 

Ve 

of  the  point's  velocity  along  the  radius  vector  is  5  ft. /sec.  Find  (a)  the 
component  perpendicular  to  the  radius  vector,  and  (6)  the  velocity  in  the 
curve. 

13.  A  crank  pin  moves  in  a  circle  2  feet  in  diameter  with  a  constant  speed 
of  28  ft. /sec.  When  the  crank  makes  an  angle  of  30°  with  the  horizontal, 
what  is  (a)  the  vertical  component  of  the  crank  pin's  velocity  ?  (6)  the 
horizontal  component  ? 

14.  The  path  of  a  projectile  is  the  parabola 


x  tan  a  — 


2  vo2  cos2  a 

and  the  horizontal  velocity  (Dtx)  is  v0  cos  a.     Show  that  the  velocity  in  the 
path  is  Vvq2  —  2  gy,  and  that  the  vertical  acceleration  is  g. 


82  THE   DIFFERENTIAL  NOTATION  [Chap.  IV. 

2  2  2 

15.  If  a  point  moves  in  the  curve  whose  equation  is  Xs  +  y*  =  c3  so  that 
the  X-component  of  the  velocity  is  constant,  find  the  acceleration  in  the 
direction  of  the  T-axis. 

16.  Find  the  equation  of  the  path  of  a  point  which  moves  in  such  a  way 
that  the  X-component  of  the  velocity  is  constant  while  the  y-component  is 
negative  and  varies  directly  as  the  time.     Give  a  physical  illustration. 

17.  Sand  or  grain,  when  poured  from  a  height  on  a  level  surface,  forms  a 
cone  with  a  circular  base  and  a  constant  angle  jS  at  the  vertex,  dependent  on 
the  material.  Let  a  denote  the  radius  of  the  base  of  the  cone  at  a  given  time, 
and  suppose  material  is  being  added  at  the  rate  of  C  cu.  ft. /sec.  At  what 
rate  is  the  radius  increasing  ? 

18.  The  velocity  of  a  jet  of  liquid  issuing  from  an  orifice  is  given  by  the 
formula,  v  =  V2  gh,  where  h  is  the  height  of  the  liquid  surface  above  the 
orifice.  If  h  =  100  feet  and  is  decreasing  at  the  rate  of  0.2  ft. /sec,  find 
the  rate  at  which  v  is  decreasing.     Take  g  =  82.2. 

19.  Given  x  =  Sy2  +  7y-\-\)  find  Dxy  without  first  obtaining  y  as  an 
explicit  function  of  x. 

20.  Air  expands  according  to  the  adiabatic  law  pv1A  =  C.  When  the 
pressure  is  40  lb.  /sq.  in.,  the  volume  is  '5  cubic  feet  and  is  increasing  at  the 
rate  of  0.2  cu.  ft. /sec.     Find  the  rate  at  which  the  pressure  is  changing. 

21.  The  formula  for  the  electrical  resistance  of  a  platinum  wire  is 

JB  =  i?o(l  +  ar  +  &r2), 
where  Bo,  «,  and  b  are  constants,  and  r  denotes  the  temperature  of  the  wire. 
Find  the  rate  of  increase  of  resistance  at  the  temperature  n,  if  the  tempera- 
ture is  rising  at  the  rate  of  0.1°  per  second. 

22.  Another  formula  for  the  variation  of  electrical  resistance  of  a  metal 
wire  with  the  temperature  is  B  =  Bo  (1  —  er  +/t2)-1.  Find  the  rate  of 
change  of  the  resistance  compared  with  that  of  the  temperature  at  tempera- 
ture T\, 

23.  Given  (x2  +  y2)2  =  a2  (x2  -  y2).  (a)  For  what  values  of  x  has  the 
curve  representing  this  function  a  turning  point!*  (b)  For  what  value  of  x 
does  the  tangent  make  an  angle  of  45°  with  the  X-axis  ?     (c)  Write  the 

equation  of  the  normal  at  x  =  -  . 

24.  A  hemispherical  bowl  20  inches  in  diameter  has  an  orifice  in  the 
bottom  through  which  the  water  contained  in  the  bowl  is  flowing.  If,  when 
the  surface  of  the  water  is  8  inches  above  the  bottom,  the  rate  of  flow  is 
16  cu.  in.  /  sec,  at  what  rate  is  the  water  level  falling,  assuming  that  the 
surface  remains  plane  ? 


CHAPTER   V 
DIFFERENTIATION   OF   TRANSCENDENTAL   FUNCTIONS 

51.  Transcendental  functions.  All  functions  that  are  not  alge- 
braic (Art.  30)  are  called  transcendental.  The  transcendental 
functions  include  trigonometric,  inverse  trigonometric,  exponential, 
and  logarithmic  functions.  In  a  former  chapter  we  considered  the 
differentiation  of  algebraic  functions;  in  this  chapter  we  shall 
develop  formulas  for  the  differentiation  of  the  more  elementary 
transcendental  functions,  starting  with  the  trigonometric  functions. 

In  the  derivation  of  the  following  formulas,  we  shall  assume  u 
to  be  a  continuous  function  of  x,  and  we  shall  find  the  derivatives 
with  respect  to  x. 

52.  Differentiation  of  sin  u. 

Let  y  =  sin  u. 

Then  y  -f  A?/  =  sin  (m  +  Aw) 

and  Ay  =  sin  (u  +  Au)  —  sin  u 

=  sin  u  cos  Au  +  sin  Au  cos  u  —  sin  u. 
Therefore,  we  have 


and 


A y  sin  u  (1  —  cos  Au)  .             sin  Au~\ 

—2-  = i '-  +  COS  U  — , 

Au  Au                                Au 

T    Ay  .          T    1  —  cos  Au  .  n           T    sin  Aul 

L    -&■  =    —  sin  u  L    - — -  +  cos  u  L . 

=  Q^u  Aw  =  0         Am                    Am  =  0    A«    J 


Au  =  oAu 
The  two  limits  in  the  second  member  have  been  evaluated ;  thus, 

L    1"cosAM  =  0,and   L    "^1       (Arts.  13,  14) 
Aw  =  0        A?*  Am~0    Au 


Hence,  L   —  =  Duy  =  cos  u. 

Au  =  OAu 


b:J 


84  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 

Since  u  is  a  function  of  xy  we  have  by  Art.  31 

Dxy  =  Duy  •  Dxu. 
Therefore  Dxy  =  cos  u  Dxu ; 

that  is,  Dx  (sin  u)  =  cos  u  Dxu.  (1) 

In  the  differential  notation,  we  have 
dy  —  cos  u  du, 
or  d  (sin  u)  =  cos  w  dw.  (2) 

In  the  special  case  where  u  =  x,  (1)  becomes 

Dx  sin  05  =  cos  a?.  (3) 

The  other  trigonometric  functions  may  now  be   differentiated 
by  applying  the  general  laws  of  differentiation. 

53.  Differentiation  of  cos  u. 

Let  y  =  cos  u  =  sin  I  ^  — 

Then  Dxy  =  cos  f  | -  w J  DJ  | -  wj 

—  —  sin  w  2>xw ; 
that  is,  Dx  (cos  u)  —  —  sin  f#  D^w,  (1) 

or,  in  the  notation  of  differentials, 

d  (cos  u)  =—  sin  u  du,  (2) 

For  m  =  x, 

D*  (cos  as)  =  -  sin  ac.  (3) 

54.  Differentiation  of  tan  u. 

Tp  .  sin  u 

If  y  =  tan  w  = , 

cos  it 


,i                                  t,         cos  ?t  ZX.  sin  u  —  sin  w  7)x  cos  u 
then  Dx?/  = x- ■ 

'  ~>SJ  u 

Dxu 


cos'  u 
cos2  w  -f-  sin2  u 


COS'  i< 

— — -  Dxu  =  sec2  w  Dxu. 
cos2  i< 


Arts.  55,  56]  DIFFERENTIATION                                           85 

Hence  Dx  (tan  u)  =  sec2  u  Dxu,                                             (1) 

or  d  (tan  u)  =  sec2  u  du.                                               (2) 

Also  Dx  (tan  oc)  =  sec2  sc.                                                   (3) 

55.  Differentiation  of  cot  u. 

-r     ,  .              COS  U 

Let  2/  =  c°t  u 

then  Dx2/ 


sin  w 
sin  u  D„  cos  u  —  cos  u  D„  sin  u 


sin2  m  +  cos2  u  t,  2     n 

= r-1- Dxii  =  —  csc^  u  Dxu. 

sm2  w 

Hence  Dx  (cot  t*)  =  -  esc2  u  Dxu,  (1) 

or  d  (cot  w)  =  —  esc2  u  du,  (2) 

For  u  =  Xj  Dx  (cot  as)  =  —  esc2  ac.  (3) 

56.   Differentiation  of  sec  u  and  esc  u. 

1 


If  2/  =  secw  = 


cos  u 


we  have  Z^?/  = Dx  (cos  w) 

cos2  u 

=  sec2  w  sin  u  Dxu  =  sec  u  tan  w  Z)xit. 

Therefore  Dx  (sec  w)  =  sec  u  tan  w  2>a,w,  (1) 

and  d  (sec  w)  =  sec  u  tan  w  rfw.  (2) 

Proceeding  as  above,  we  find 

2>aj  (CSC  U)  =  —  CSC  w  cot  w  Dxu9  (3) 

or  <Z  (esc  u)  =  —  esc  w  cot  w  du.  (4) 

The  details  of  the  proof  are  left  to  the  student. 

Ex.     Differentiate  f(x)  =  tan2  Va2  -  x2. 


We  have  y  =  tan2  Va2  —  x2, 


whence  J5xy  =  2  tan  Va2  -  x2  Z)x  tan  Va2  —  x2. 


86  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 


From  (1)  Art.  54,  taking  u  =  Va2  —  x2,  we  obtain 


Z>ztan  Va2  —  x2  =  sec2  Va2  —  x2  Dx  Va2  —  x2 

=  -  sec2  Va2-x2 

Va2  -  x2 

Hence,  /'(x)  = —  —  tan  Va2  -  x2  sec2  Va2  -  x2. 

Va2  —  x2 

EXERCISES 

Differentiate  the  following. 

1.   y  =  cos  ax.  2.   y  =  tan8  x. 

3.   y  =  sin  2  x  —  cos  2  x.  4.   y  =  sec2  x. 

5.   y  =  sin2  3  x.  6.    y  =  cos3  (a2  —  x2). 

7.    y  =  x  sin  x.  8.    ?/  =  x2  tan  -  • 

— 


9.   ?/  =  cos  vx2  —  a2.  10.    y  =  x  —  tan  x. 

11.   2/  =  x3csc2x.  12.   y  =  sinx  —  xcosx. 

13.  y  =  2  x  cos  x  +  (x2  -  2)  sin  x.  14.  y  =  x  —  sin  x  cos  x. 

15.  y  =  sin  x  (cos2  x  +  2).  16.  p  =  2a  — • 

y  cos  0 

i 


17.    p  =  a  Vcos  2  0.  18.  p  =  a  (sin  n0) 

19.  /o  =  asec3|-  20.  />  = 


cos0 


21.  s==  Acoaut  —  B  sin  w£. 

22.  s=-  r  cos  0 


+  7^1-AJl-gsin2^. 


23.  Find  the  derivative  Dxy,  when 

x  =  a(0  —  sin0), 
y  =  a(l  -  cos0). 

24.  Let  x  =  a  cos  0  +  «0  sin  0, 

?/  =  a  sin  0  —  a0  cos  0. 
Find  Ztyx,  Z>^^,  and  Dxy. 

25.  Find  the  polar  subtangent  and  polar  subnormal  of  the  curves  : 
(a)  />  =  sin0.  (6)  />  =  a(l  —  cos  0).  (c)  /o  =  asec2-- 

26.  From  the  equation  sin  2  0  =  2  sin  0  cos  0,  derive  by  differentiation  0 
formula  for  cos  2  0. 

27.  Find  expressions  for  the  subtangent  and  subnormal  of  the  curve 
y  —  a  sin  x. 


Art.  57] 


INVERSE   TRIGONOMETRIC   FUNCTIONS 


87 


28.  Find  the  angle  at  which  the  curves  y  =  cos  x  and  y  =  tan  x  intersect. 

29.  Find  the  angle  which  the  curves  y  =  sin  x  and  y  =  cos  x  make  with 
each  other  at  their  points  of  intersection. 

30.  Find  the  value  of  0  for  which  tan  0  is  in- 
creasing twice  as  fast  as  0. 

31.  When  0  =  22u,  find  approximately  the 
changes  in  sin  0  and  cos  0,  for  a  change  of  1'  in 
the  angle. 

32.  In  a  triangle  two  sides  a  and  b  include 
an  angle  0.  If  the  sides  remain  of  constant 
length  and  the  angle  is  varied,  find  the  rate  of 
change  of  the  area  of  the  triangle  with  respect  to 

0  when  0  =  -- 
4 

33.  In  a  certain  type  of  motion  the  velocity 
is  given  by  the  expression  v  =  v0  cos  kt.  Find  an 
expression  for  the  acceleration. 

57.    Inverse  trigonometric  functions.    The 

trigonometric  functions  are  all  single- 
valued  functions ;  that  is,  for  each  value  of 
the  variable  there  corresponds  one  and 
only  one  value  of  the  function.  Thus,  from 
x  =  sin  y,  x  has  but  one  value  for  each 
value  of  y.  The  inverse  trigonometric 
functions  are  not  single-valued,  but  mul- 
tiple-valued functions,  for  to  each  value 
of  variable  there  corresponds  any  number 
of  values  of  the  function.  In  Fig.  22  is  given  the  graph  of 
y  =  arc  sin  x.  For  x  =  a,  it  will  be  seen  that  y  may  have  any  one 
of  the  values  indicated  by  the  points  Mx,  M2,  M3,  •••.     If,  however, 

the  values  of  y  be  restricted  to  the  interval  [  —  ^,  ^ ),  then  within 

this  range  the  function  is  single-valued. 

The  graph  of  y  =  arc  sec  x  is  given  in  Fig.  23.  From  the 
examination  of  this  graph,  it  will  be  seen  that  the  function  will 
be  single- valued  if  0  ^  y  <£  it. 

Ex.  Plot  roughly  the  graphs  of  arc  cos  x,  arc  tan  cc,  and  arc  cot  x,  and 
determine  limits  within  which  these  functions  are  single-valued. 


88 


TRANSCENDENTAL  FUNCTIONS 


[Chap.  V. 


I 

tit 

<z 

>J 

7T 

7T 
2 

r~ 

0 

-/^ 

> 

-7T 

Fig.  23. 


58.    Differentiation  of  arc  sin  u  and  arc  cos  u. 

Let  y  =  arc  sin  w, 

then  u  —  sin  ?/. 

Differentiating,  we  obtain    Dyu  =  cos  y. 

1 


Therefore 
and 

Since 
we  have 


Duy  = 

cosy 

Dj)  =  Duy>Dxu  (Art.  31) 

=  J—Dxu. 

cosy 

cos  y  =  Vl  —  sin2  ?/  =  Vl  —  w2, 
1 


Al 


vr^~ 


ZU. 


Arts. 

68, 

59] 

DIFFERENTIATION 

Henc 

e, 

Dx  arc  sin  u  = 

1 

Vl  -  u* 

or  in 

the  differential  notation, 

d  arc  sin  u  = 

du 

Vl-^2 

If, 

as 

a  special 

case,  u  =  x, 
Dx  arc  sin  oc  = 

1 

89 

Z>xU,  (1) 

(2) 

(3) 

Proceeding  in  the  same  way  with  y  =  arc  cos  u,  we  obtain 

Da-  arc  cos  t*  = ==  D^w,  (4) 

dare  cos  w= ^*__,  (5) 

Da,  arc  cos  a?  = .  (6) 

Vl-^2 


Since  the  sign  of  the  radical  Vl  —  x2  may  be  either  positive  or 
negative,  the  signs  of  the  derivatives  in  (3)  and  (6)  are  ambiguous. 
Reference  to  the  graph  of  arc  sin  x,  Fig.  22,  shows  how  the  ambi- 
guity arises.  Thus  for  x  =  a,  the  derivative  is  positive  at  Mlf  Ms, 
M5,  etc.,  and  negative  at  points  M2,  Mi}  etc.    When  y  is  restricted 

to  the  interval  —  ^  <  y  <  ^  so  as  to  make  the  function  single- 

valued,  the  radical  must  be  taken  with  the  positive  sign.  The 
student  may  show  that  if  0  fg  y  ^  7r,  the  function  y  =  arc  cos  x  is 
single-valued  and  the  derivative  given  by  (6)  has  the  proper  sign 
when  the  radical  is  given  the  positive  sign. 

59.    Differentiation  of  arc  tan  u  and  arc  cot  u. 

Let  y  =  arc  tan  u, 

whence  u  =  tan  y, 

and  Dyxi  =  sec2  y  =  1  +  u2. 

1  1 


We  have  then  Duy  = 

But  by  Art.  31,  Dxy  =  Duy  •  Dxu  =  ^—-2  Dxu ; 


sec2  y     1  +  u2 

1  +  "2 


90  TRANSCENDENTAL   FUNCTIONS  [Chap.  V. 

therefore,  !>«.  arc  tan  u  =  — - — -  Dxu,  „        (1) 

1  -\-u* 

and  d  arc  tan  u  =    du  - .  (2) 

1  + 1*2  w 

For  u  =  x  Dx  arc  tan  x  =  — ^— .  (3) 

1  +  x2 

To  differentiate  arc  cot  u,  we  may  proceed  as  above,  or  we  may 
derive  the  result  from  (1).     Thus,  since 

y  =  arc  cot  u  =  arc  tan  - , 
u 


D'y = Jlffi  °"  © = ^T1? '  (~  ^D'W =~r^?  D*U' 


Hence,  we  have     Dx  arc  cot  u  = — -  D^U,  (4) 

1  +  U* 


du 

1  + 


d  arc  cot  W=-_^M_,  (5) 


2>a,  arc  cot  a?  = s .  (6) 


60.  Differentiation  of  arc  sec  u  and  arc  esc  u. 

Let  .       y  =  arc  sec  u, 

then  w  =  sec  y, 

and  Z)yw  =  sec  y  tan  y. 

Hence,  we  have  Duy  — , 

sec  y  tan  y 

and  Dxy  =  Duy  Dxu  (Art.  31) 

1 


sec  y  tan  y 


Dju. 


Since            sec  y  =  uf  and  tan  ?/  =  Vsec2  y  —  1  =  Vw2  —  1, 
we  may  write  Dxy  = r=  Z>,w. 

w  Vw2  —  l 


Art.  60] 

DIFFERENTIATION 

Therefore 

u  VMa  _  l 

End 

rf  nrp  hoc  11  —         (^u 

u  Vwa  _  i 

Also 

ac  Va?2  -  1 

Proceeding  in  the  same  way,  we  obtain 


91 

nxu9  (l) 

(2) 
(3) 


Dx  arc  esc  u= Ac**,  (4) 

u  Vw2  _  i 

d  arc  esc  u  = du       ,  (5) 

u  Vw2  _  i 

Dx  arc  esc  a?  = .  (6) 

a?  v^  -  1 

From  the  graph  of  arc  sec  jd,  Fig.  23,  it  is  readily  seen  that  for 
0^y  =  -,  the  derivative  is  positive  and  therefore  the  positive 

a 

sign  of  the  radical  V#2  —  1  must  be  taken ;  but  for  --^.y^ir  the 

A 

derivative  is  positive  while  x  is  negative,  and  therefore  for  this 
interval  the  negative  sign  of  the  radical  must  be  taken.  From 
the  graph  for  arc  esc  x  the  student  may  deduce  a  similar  statement 
relative  to  the  radical  in  (6). 

Ex.    y  =  arc  tan 


VI  -  2  x2 

du 


Let  —  =  u  ;  then  y  =  arc  tan  u,  dy  = 


VI  -  2  x*  l+u 

VI  -  2  a?  + 


Vl  -  2  a*  1  _ 

and      du  =  d — ==  == ^ jr— 5 «£  = -  dx. 

Vl  -  2  *2  1  -  2  a*  p  _  2  ^1 

Substituting  these  values  of  u  and  du,  we  get 

1  1  1  j 

(fa  = 5 r  efcc  = ax. 

1+  *  (1-2*2)4  (1-*2)V1-2X2 

1  -  2  x*     K  } 


92  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 

EXERCISES 

Differentiate  the  following : 

1.   y  =  arc  cos  -  •  2.   y  =  arc  tan  -^-  • 


Vk 


r'2 


3.    y  =  arc  sec  2L  .  4.   y  -  arc  sin  Va2  —  x2. 


5.   y  =  x  arc  tan  jc.  6.0  =  arc  sec 


Vp2  -  a2 


1.6  =  arc  sin  ^ ^_  •  8.    y  =  arc  tan  -  +  arc  tan 

P 


a 

x 


9.   y  =  arc  cos  (cos  x).  10.  1/  =  arc  tan  VI  —  Jc2  tan2  x. 

1  cc 

11.   y  =  x*  arc  cos  x2.  12.    y  =  —  arc  tan  _ . 


Va2  —  x2  ,  .    x 


13.   ?/  = h  arc  sin  -  •  14.    ?/  =  Vx2  —  a'2  —  a  arc  sec  - 

x  a  a 

15.    ?/  =  -  -  Va2  -  x2  +  —  arc  sin  -  . 
2  2  a 


16.    y  =  x  arc  cos  x  —  Vl  —  x2. 

61.  Exponential  and  logarithmic  functions.  The  differentiation 
of  the  exponential  functions  au  and  eu  depends  upon  the  evalua- 
tion of  the  limit  ax  —  1 

L  > 

x  =  0      % 

and  this  limit  in  turn  involves  the  limit  L  ( 1  +-  j  .      It  will  be 


x) 

necessary  to  consider  these  two  limits  before  the  desired  formu- 
las for  differentiation  can  be  deduced. 

Jj    ( 1  H —  )   .     It  can  readily  be  shown  that  L    ( 1  +  -  )  ,  where  n  is  a 

positive  integer,  is  some  number  lying  between  2  and  4.  The  outline  of  the 
proof  is  as  follows  :  Take  two  numbers  a  and  b  such  that  a  >  b  >  0  and 
let  n  be  a  positive  integer.     Then  we  have  the  identity, 

qn+l-frn+l  _  ^  +  ^^  +  ^_^  +       #  +  (1) 

a  —  6 
from  which  follows  the  inequality 

a  —  0 
or  fl»+i  -  &1+1  <  aw(a  -  b)  (n  +  1). 


Art.  61]    EXPONENTIAL  AND   LOGARITHMIC   FUNCTIONS  93 

The  last  inequality  may  be  thrown  into  the  form, 

an[a  -  (a  -  6)  (n  +  1)  ]  <  bn+\  (2) 

Now  choose  for  a  and  b   two  sets  of  numbers,    subject  to  the  condition 
a  >  b  >  0.     Let  these  be  : 

(a)     a=l+l,  &  =  1  +  _1_, 
n  n  +  1 

(6)     o  =  l  +  ~,  5  =  1. 

2  n 

Substituting  these  values  respectively  in  (2),  we  get  the  two  inequalities 

<  I-  (4) 


2  V     2  wy 


From  (3)  it  appears  that  the  function  (1+-)    increases  with  n  ;  that  is, 

the  function  is  monotone.     For  n  =  1,  the  function  takes  the  value  2.     From 
the  inequality  (4)  we  obtain,  by  squaring  both  members, 


K-T 


<4, 

whence  it  follows  that  the  function  f  1  +  -  )     cannot  exceed  4. 


The  limit  required  must  therefore  exist  (Art.  14)  and  it  lies  between  2  and 
4.  The  exact  value  of  this  limit  is  2.7182818285  •••,  as  will  be  shown  here- 
after in  connection  with  infinite  series.  This  number  is  denoted  by  e  and  is 
the  base  of  the  natural  system  of  logarithms. 

In  the  preceding  discussion  n  was  given  only  positive  integral  values.  The 
limit  is  the  same,  however,  if  instead  of  n  we  take  a  variable  x  which  can 
take  all  real  values.     Hence,  in  general,  we  have 


L 

*  =  ae 


(l  +  i)"  =  e.  (5) 


fiX   1  ~x  1 

L    .     To  obtain  the  limit  of    as  x  approaches  zero,  let 

a?  =  0       35  X 

=  ax  —  1.     Then  x  =  \oga  (1  +  y)  and  as  as  =  0,  y  =  0.     We  have  therefore 
L    *L=±=  L   V =  jr    1 

X  =  Q       X  y=0log«(l+*/)        y  =  0  1 

loga(l  +  y)j, 
1  1 


-1        l0ga 
l0ga    L     (l+|f)» 
y  =  0 


94  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 

Since,  however, =  log  a.  we  have  finally 

loga  e 

OK  -t 

log  a.  (6) 


se  =  0       SB 

The  substitution  of  e  for  a  in  (6)  gives 


L    ^— =-l=-loge  =  l.  (7) 


62.  Differentiation  of  «**  and  eu. 

Let  ?/  =  aM ; 

then  y  +  Ay  =  aM+A", 

and  Ay  =  aM+A"  —  aM  =  aM  (aAM  —  1). 


Ay        uaL 


Therefore  —±=za; 

Au  Au 

Taking  limits,  we  have 

Au±0&u        Aw  =  0     a% 

But,  as  shown  in  the  preceding  discussion, 

T    aAM-l      . 

L    =  log  a  ; 

Aw  =  0     Aw 

hence,  we  have  Duy  =     L    :_£  =  aM  log  a. 

Aw  =  oAw 

Using  the  theorem  of  Art.  31,  we  have 

Dxy=Duy-Dxu 
=au  log  a  Dxu\ 
that  is,  2>x<*>u  =  <*>u  log  a  Dxut  (1) 

or  daw  =  a11  log  a  d u.  (2) 

If  u  =  x,  we  have 

D^a*  =  a*  log  a,    or    dax  =  ax  log  a  dx.  (3) 

If  in  (1)  we  substitute  e  for  a,  we  obtain  (since  log  e  =  1) 

nxeu  =  eunxu;    deu  =  e1ldu.  (4) 


Arts.  63,  04]  DIFFERENTIATION  95 

For  u  =  x,  we  have 

jD^e*  =  e* ;  dex  =  ex  due.  (5) 

It   may  be  noted  that  ex  is   a  function  whose  derivative  is 
the  function  itself. 

63.    Differentiation  of  loga  u. 

Let  y  =  loga  u, 

whence  u  =  ay. 

From  the  preceding  article,  we  obtain 

Dyu  =  ay  log  a, 

whence  Duy  = 


ay  log  a 
By  the  theorem  of  Art.  31, 

Dxy  =  Duy  •  Dxu 

=  — ; Dxu  =  — Dxu. 

ay  log  a  u  log  a 

We  have  therefore 

na>logau  =  -L-2»!it  or  dlogau  =  -±-*».  (1) 

loga    u    '  loga  u  v  J 

If  u  =  x,  we  have 

Z>*logaK=rJ—^;  dlogaac  =  -^—^.  (2) 

log  am  loga  ac 

The  fraction  is  called  the  modulus  of  the  system  of  log- 
log  a 

arithms  whose  base  is  a,  and  may  be  denoted  by  m.     For  the 

Briggs'  or  common  system,  in  which  a  =  10,  m  =  .434294  ••♦. 

We  have,  therefore,  m 

Dxlogax  =  -. 
x 

64.   Differentiation  of  log  u. 

By  the  substitution  of  e  for  a  in  (1)  and  (2)  of  the  preceding 
article,  the  following  formulas  are  obtained  (since  log  e  =  1)  : 

nxlogu  =  -D:>l,u;    dlogu  =  —.  (1) 

u  u 

I>xlogx  =  ±;  dlogoc  =  dv.  (2) 

90  HC 


96  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 

Ex.  1.  y  =  log  sin  x. 

dy  — d  sin  x  = cos  x  dx  =  cot  x  dx. 

sin  x  sm  x 

Ex.   2.  y  =  eVi+*\ 

dy  =  e^^+J  <Z(Vl  +  x2)  =  ev/i+^       x    -  dx. 

VI  +  x2 

65.  Logarithmic  differentiation.  If  a  function  consists  of  a 
number  of  factors,  it  may  be  conveniently  differentiated  by  tak- 
ing logarithms.     Thus,  if 

y  =  uvw, 

log  y  —  log  u  -f  log  v  -f-  log  to. 

Differentiating,  we  have 

dy  _  du     dv     dw 
—        I         I        ) 
y        2i        v        iv 

or  dy  =  *-  du  +  ^  dv  +  -^  c?w. 


Ex.  1.     Differentiate  the  function  y 


Vx2  -  8 


(x  +  7)2 
Taking  logarithms,  we  have 

log  y  =  log  *  +  I  log  (x2  -  3)  -  2  log  (x  +  7), 
whence  by  differentiation 

«&  =  (!  +  -* LU 

y      \x     x2-3     x+1) 

=  14x2  +  3x-21  g 
*  (x  +  7)  (a;2  -  3) 
Therefore 

dy  _xVx2-S      14  a;2  +  3  x  -  21 
dx      (x  +  7)2    '  x(x  +  7)(x2  -  3) 

_  14  x2  +  3  x  -  21 

(x  +  7)8Vx2~^3' 

Logarithmic  differentiation  is  especially  useful  in  differentiating 
an  exponential  function  having  a  variable  base. 


Akt.  65]  LOGARITHMIC   DIFFERENTIATION  97 

Ex.  2.   Given  y  =  exxx, 

we  have  log  y  =  log  ex  +  -  log  x 

x 

=  x  +  -log  a;. 
x 
Differentiating,  we  have 

JL  —  dx log  x  dx  +  —  dx, 


dy 
dx 


1  1-2 

e*xx  +  exxx    (1  —  logx). 


Logarithmic  differentiation  also  enables  us  to  differentiate  un, 
where  n  may  have  any  constant  value.     We  have 

y  =  un, 

whence  log  y  =  n  log  u. 

Differentiating,  we  obtain 

ln  An  v 

-Dxy  =  -Dxu, 
V  u 

whence  Dxy  =  n  ^  Dxu. 

u 

or  Dxy  =  nun  ~  1I>gBu. 

This  completes  the  demonstration  begun  in  Art.  29,  and  shows 
that  the  formula  given  there  for  the  derivative  of  un  holds  for  all 
real  values  of  n. 

EXERCISES 

Differentiate  the  following : 

1.    2/  =  log(*2_3x  +  5).  2.    y  =  lQg7*  +  5»y 

\x  —  m) 

4.    y  =  e*3  +  ex. 

6.   |f  =  log  (log*). 

8.   y  =  «x  +  e-x. 
ex  —  ex 

10.    y  =  log  tan  x. 


3. 

j,  =  a^xJ^\ 

5. 

y  =  x\ogx. 

7. 

y  =  ex  +  e~x. 

9. 

y  —  gsin*. 

98  TRANSCENDENTAL  FUNCTIONS  [Chap.  V. 


Vx2  +  1  - 


11.   y  =  log  v  _ ^   — -•  12.   y  =  log(x  +  vx2-a2). 
vx2  +  1  +  x 

i  x  . 

13.   y  =  -log ,  14.    y  =  x arc  tan  x  —  log  V 1  +  x2. 

a       a+  vx2  +  a2 

15.   y=^(ax-l).  16.   ^-A-log^6*" 


26    "      b 


In  Exs.  17  to  20,  differentiate  by  taking  logarithms 


17.    y  =  x(l  -  x)  Vl  +  x2.  18.   y  = 


x2  Vx2  —  4 


19.,  =  ^^-  20.   ,  =  XS 

XV  1  —  X2  *1  -fx 

21.  If  z  =  Ae^'k  +  Be-*/*,  find  —  . 

22.  Find  an  expression  for  the  velocity  when  the  distance  traversed  by  a 
moving  point  is  expressed  as  a  function  of  the  time  by  the  equation 

s=(A  +  Bt)e~kt. 

23.  If  s  =  e~kt  [a  sin  mt  +  b  cos  mt~] ,  find  an  expression  for  the  velocity. 

24.  Find  the  polar  subtangent,  polar  subnormal,  lengths  of  polar  tangent 
and  polar  normal  of  the  curve  p  =  eaQ. 

25.  Find  the  angle  between  the  curve  y  =  log  x  and  the  axis  of  x  ;  between 
the  same  curve  and  the  line  y  =  2. 

26.  Find  the  subtangent,  subnormal,  and  the  length  of  the  normal  of  the 

catenary  y  =  «(e«  +  e~«). 

2 

27.  Given  log 4.32  =  1.4633,  find  approximately  the  value  of  log  4.33  by 
means  of  the  theorem  Ay  =/'(x)Ax. 

28.  By  logarithmic  differentiation  derive  the  result  dau=au  log  a  du. 


MISCELLANEOUS   EXERCISES 

1.  Derive  the  formulas  for  the  derivatives  of  cos  w,  tan  u,  and  sec  u 
directly  from  the  defining  equation 

jr     A?/  _    L    f(u  +  Au)-f(u) 

and  the  theorem  of  Art.  31. 

2.  Obtain  the  derivative  of  cot  u  from  the  relation  cot  u  = :  also 


the  derivative  of  sec  u  from  sec  u  =  Vl  +  tan2  ?«. 


tan  u 


Art.  65]  MISCELLANEOUS   EXERCISES  99 

3.  Write  out  formulas  for  the  derivatives  of  the  inverse  circular  functions 

for  the  case  in  which  u  =  -  • 
a 

4.  Discuss  the  following  curves  by  means  of  their  derivatives.  Find 
where  they  cross  the  X-axis,  the  angles  at  which  they  cross,  and  the  point  at 
which  their  tangents  are  parallel  to  the  X-axis. 

(a)  y  =  sin  (  x  —  -  V  (&)  y  =  sin2a:. 

(c)  y  =  sin  x  +  cos  x.  (d)  y  =  log  x. 

5.  When  0  =  36°,  what  is  the  rate  of  increase  of  (a)  sin  0 ;  (6)  cos  0 ; 
(c)  tan  0  compared  with  the  rate  of  increase  of  0  ? 

6.  The  cycloid  is  given  by  the  equations 

x  =  a{0  —  sin0), 
y  —  a{\  —  cos0). 

Derive  an  expression  for  the  slope  of  the  tangent,  and  find  the  value  of  this 
slope  when  a  =  3,  6  =  f  7r.  For  what  value  of  0  is  the  tangent  parallel  to  the 
X-axis  ? 

Differentiate  the  following : 


7.  v^r^-aiog«  +  v<*2-*2. 

X 

o    x  —  8  a 


V2  ax  +  a;2  +  -  a2log  (as  +  a  +  V2  ax  +  x2). 

-  A 


9.   ^V«2-z2+^arcsin-. 
2  2a 

1        1  - 

10. 1-  arc  tan  x.  11.   log(e*  +  e~x).  12.    tanax. 

x      '3  Xs 

13 .  Find  the  polar  subtangents  and  subnormals  of  the  following  curves : 

(a)  p  =  a(l  +  cos0).  (6)  p2  =  ea». 

0 
(c)  />  =  a2sin20.  (d)  p  =  sec22* 

14.  Find  expressions  for  tan  f  (Art.  40)  for  the  curves  of  Ex.  13. 

15.  If  a  point  moves  in  a  logarithmic  spiral  p  =  eae,  show  that  the  compo- 
nents of  its  velocity  along  and  perpendicular  to  the  radius  vector  have  a  con- 
stant ratio. 

16.  A  point  moves  in  the  curve  p  =  a(l  —  cos  0)  with  a  speed  of  m  units. 
Find  the  velocity  components  along  and  perpendicular  to  the  radius  vector. 


100 


TRANSCENDENTAL   FUNCTIONS 


[Chap.  V. 


17.   The  following  formulas  give  approximately  the  relation  between  the 
pressure  and  temperature  of  saturated  steam  : 


(a)  logi>  =  ^  +  |+^2- 
(6)  log  p  =  m  —  n  log 


(T=r  +461) 


T-  C 

(c)  logjo  =  a  +  ha*  -  c/3«.  (0  =  r  -  32) 

From  each  formula  derive  an  expression  for  the  rate  of  change  of  the  pres- 
sure with  the  temperature. 

18.   A  point  moves  in  accordance  with  the  law  expressed  by  the  equation 

s  =  ae~Kt  cos  2  ir(bt  +  c). 
Derive  an  expression  for  the  velocity. 

19.  A  radius  OP  of  length  a,  Fig.  24, 
rotates  with  constant  angular  speed  w  about 
.  y-  the  point  0,  and  the  projection  M  of  P  on 
the  X-axis  therefore  moves  on  this  axis. 
Show  that  OM  =  x  =  a  cos  <at,  and  derive 
expressions  for  the  velocity  and  acceleration 
of  the  point  M. 

F      24  (The  motion  of  M  is  called  simple  har- 

monic motion.') 

20.   Taking  the  time  t  as  abscissa,  draw  curves  showing  the  displacement 
x,  velocity  v,  and  acceleration  a  of  the  point  M.     Take  t  =  0  when  M  is  at  A. 


CHAPTER   VI 
INTEGRATION 

66.  Anti-derivatives  and  integrals.  Thus  far  we  have  been 
chiefly  concerned  in  finding  the  derivatives  of  given  functions. 
We  shall  now  consider  the  inverse  operation;  that  is,  having 
given  a  function  <f>(x),  to  find  another  function  f(x)  such  that 
Dxf(x)  =  <f>(x).  This  inverse  operation  is  called  integration,  and 
the  resulting  function  is  called  the  anti-derivative  or  integral  of  the 
given  function.     The  function  integrated  is  called  the  integrand. 

Two  symbols  are  in  use  to  denote  the  inverse  operation  of 
integration.  The  symbol  D'1  may  be  prefixed  to  the  integrand ; 
thus,  since 

Dxvi  =  3x2, 
we  have 

i).-13aj2  =  aj». 

It  is  the  universal  custom,  however,  to  denote  integration  by  plac- 
ing the  symbol    I   before  the  differential.     Thus,  since 

d(a?)=3a?dx, 
we  write  J  3  x2  dx=x3. 

The  direct  and  inverse  operations  may  therefore  be  indicated  by 
the  symbols  Dx  and  D~l ;  or  by  the  symbols  d  and   I  . 

EXERCISES 
Find  anti-derivatives  of  the  following  functions  : 
1.   5x*.  2.   2x.  3.   cosz.  4.    1-  5.    --• 

X  X2 

Perform  the  following  integrations : 

6.     fsec20d0.  i    7.    (2atdt.  8.    $ ^Itf'  9*   ^ dx' 

101 


102  INTEGRATION  [Chap.  VI. 

67.  General  theorems.  It  will  be  observed  that  the  functions 
Xs  and  Xs  -J-  C,  where  C  is  any  constant,  have  the  same  differential 

3  x2  dx.     Hence  the  integral    I  3  x2  dx   should   have   the   general 

form  x*  +  C.  In  general,  since  the  derivative  of  a  function  plus 
a  constant  is  always  equal  to  the  derivative  of  the  function,  we 
have  the  following : 

Theorem  I.  The  general  form  of  the  anti-derivative  or  integral 
of  a  function  must  involve  an  additive  constant.  This  constant  is 
arbitrary,  and  is  called  the  constant  of  integration. 

The  value  of  the  constant  of  integration  in  any  particular  case 
may  be  determined  when  certain  initial  conditions  in  the  problem 
under  discussion  are  known.  The  determination  of  this  constant 
from  such  conditions  will  be  subsequently  discussed  more  fully. 

An  anti-derivative  or  integral  function  may  be  represented  by  a 
graph.  The  geometrical  significance  of  annexing  the  constant  of 
integration  is  to  increase  or  decrease  each  ordinate  of  this  graph 
by  the  same  length.  In  other  words,  as  we  give  various  values  to 
the  constant  of  integration,  we  move  the  integral  curve  up  or 
down.     See  Fig.  8. 

It  has  been  shown  that  the  derivative  of  the  product  of  a  con- 
stant and  a  function  is  the  product  of  the  constant  and  the  de- 
rivative of  the  function ;  that  is,  Dx  c  -f(x)  =  c  •  Dxf(x)  or  in  the 
differential  notation,  d(cu)  =  c  du.  The  inverse  operation  gives  a 
similar  theorem,  namely : 

Theorem  II.     A  constant  factor  in  the  integrand  may  be  ivritten 

either  before  or  after  the  symbol  i  ;  that  is 

\cdu=c\du.  (1) 

The  derivative  of  the  sum  of  a  finite  number  of  functions  was 
obtained  by  adding  the  derivatives  of  the  separate  functions. 
The  inverse  operation  gives,  therefore,  the  following  theorem : 

Theorem  III.  The  integral  of  the  algebraic  sum  of  a  finite  num- 
ber of  functions  is  equal  to  the  algebraic  sum  of  the  integrals  of  these 
functions;  that  is, 

((du  +  dv  +  dw)  =  (du  +  (dv+\dw.  (2) 


Art.  68]  THE   INTEGRAL  103 

.    The  integral   J  utl  du.     By  differentiating  the  function  un+l, 
d(un+1)  =  (n  +  l)un  du, 


68 

we  obtain 


whence  dl ]  =  un  du. 


J1  +  h 
Therefore,  we  have 

\undu=^ — -+C, 
J  n+1 

which  holds  for  all  values  of  n  except  the  value  —  1.  This  for- 
mula is  especially  useful  in  the  integration  of  integral  algebraic 
functions. 

Ex.   1.      (5xBdx  =  5(x*dx  =  %x*+C. 

Ex.    2.      f  (4  x2 -  3)3x dx.     Put  w  =  4  x2  -  3,  whence  du  =  8  x dx.    The 

integral  then  takes  the  form  £  (  u9  du.     Hence 

f(4x2-3)3xdx  =  i  (4 x2  -  3)4  +  C. 

EXERCISES 

Verify  the  following : 

1.     f x4  dx-  =  \x*>+  C.  2.     (x%  dx  =  §  x*  +  0. 

3.  f  (x3-2x  +  5)a*x  =  £x4-x2  +  ox  +  c. 

4.  f(3»-2-5x-4)dx=-3x-1  +  fa;-3+  C. 
Evaluate  the  following  integrals  : 

5.  f  (ax~3  +  bx-*)  dx.  6.    f  (5  x6  -  4  x3  +  2  x  +  7)  dx. 

7.  J(3ar*  +  4-a£  +  2s*)cfc;.  8.  J^l+I  +  jW 

9.  j*(»*- 8 «*)<*&  10.     f(2x2-3x)4«x. 

11.  §(ax  +  b)*dx.  12.     (*3  x2(x3+  4)2ax. 

13.  j'xX2x4-5)3^.  14.     f  x(3  x2-  7)4  dx. 

15      r  (3a;2-5)(la;  16.     f_*dx_. 

J  Vx3  -  5  x  +  7  ^(x2  -  5)* 


104  INTEGRATION  [Chap.  VI. 

69.  Fundamental  integrals.  From  the  results  obtained  in  the 
preceding  chapters  we  have  the  following  list  of  fundamental 
integrals.  The  student  should  make  himself  thoroughly  familiar 
with  these  formulas. 

C  attl+l 

1.    \undu  =  - — -+C.     (w=£-l). 
J  n  +  \ 

2.   f  cos  u  du  -  sin  u  +  C.  3.  J  sin  u  du  =  -  cos  u  +  C. 

4.   f  sec2  u  du  =  tan  u  +  C.  5.  J  esc2  udu  =  -  cot  te  +  C. 

6.   f  sec u  tan  w  dw  =  sec u  +  C  7.  J  esc ucotudu=  —  esc  w  +  C. 

8.   f     d™     =arcsint4  +  C  9.  f-^-  =  arctanw+ C 

=  -arc  cos  w  4-  C.  =-  arc  cot  m  +  C. 

10.   f du       =  arc  sec  m  +  C       11.   f  «m<Im  =  =-^-  +  C 

J^vW>_i  J  log« 

=  —  arc  esc u+C. 

12.   f eu  du  =  eu+  C.  13.   f^  =  log  «*  +  C. 

The  following  additional  integrals  are  important  and  should 
also  be  committed  to  memory. 

14.    f_^=    1   l0g^«  +  O.     (**>«*) 

=  J_  log«^+C.    (**<«*). 
2«        a  +  u 

is.  f     dM     =  iflp(tt4-Vtt» +  «*)+<?■ 


16.  f  tan  udu-  log  sec  m  +  C 

17.  (cot  i*  fit*  =  log  sin  u  +  C, 

18.  f  esc  u  du  =  log  tan  ~+  C, 

19.  J  sec  w  du  =  log  tan  (|  +  ~\  +  C. 


Art.  70]  INTEGRATION  BY  INSPECTION  105 

The  derivation  of  these  last  forms  will  be  given  later.  The 
student  may  here  verify  them  by  differentiation. 

70.  Integration  by  inspection.  The  integrals  tabulated  in  the 
preceding  article  are  called  fundamental  or  standard  integrals.  In 
these,  the  integrands  are  recognized  as  the  results  of  previous 
differentiations;  hence,  if  a  given  integrand  has  one  of  these 
forms,  the  integral  is  known  at  once  as  the  function  previously 
differentiated.  When  the  integrand  has  not  one  of  these  standard 
forms,  it  must  be  prepared  for  integration;  that  is,  by  suitable 
transformations,  it  must  be  reduced  if  possible  to  an  integrand  of 
standard  form  or  to  a  sum  of  such  integrands.  The  general  method 
used  for  such  reductions  forms  the  subject  matter  of  Chapter  XII. 
In  the  present  chapter  we  shall  consider  only  such  reductions  as 
are  quite  obvious. 

It  should  be  noted  that  the  variable  u  in  the  standard  forms 
may  be  any  function,  /(a*).     For  example,  consider  the  integral 


/< 


e8inxcos#c?sc. 
Since  cos  xdx  =  d  (sin  x),  this  may  be  written 

esinxd  (sin  x), 


S' 


which  will  be  recognized  as  form  12,  with  u  replaced  by  sin  x. 

Frequently  the  integrand  may  be  made  to  assume  a  standard 
form  by  the  introduction  of  a  constant  factor.  The  following 
examples  are  illustrative : 

Ex.1.  f(3a;-2)2{to.     ■ 

Here  the  standard  form  (undu,  where  w  =  3x-2,  and  »  =  2,  is  sug- 
gested. Since  du  =  d(3x-  2)=  8  dx,  the  reduction  is  effected  by  introduc- 
ing the  factor  3  in  the  integrand  and  the  neutralizing  factor  \  outside  of  the 
integral  sign.     Thus,  we  have 

((Sx-2ydx  =  l((3x-2y3dx=±$(3x-2yd(3x-2)=%(3x-2y+C. 

Ex.  2.      (eax+bdx=-  (ea*+badx  =  -(eax+bd(ax+b)  =  ~— —  +  C. 
J  aJ  aJ  a 


106  .  INTEGRATION  [Chap.  VI. 

Here  the  introduction  of  the  factor  a  reduces  the  integrand  to  form  13,  u  . 
being  replaced  by  ax  +  b. 

Ex.  3.  f     xdx      =  ( (a  -  bx2)~h dx. 

J  Va  —  bx2     J 

Introducing  the  factor  —  2  b,  we  get 

_  JL  f  (a  _  bx2)~\-2  bxdx)  =  -^-  f  (a  -  bx*)~*  d(a  -  bx2). 
2  o  J  «  bJ 

This  integral  will  be  recognized  as  the  standard  form  1  with  u  =  a  —  bx2, 
and  n  =  —  \.    The  result  is  therefore 

__L.  (a-Wi+1+C  =  -±(a-b&)l+C. 
26  -J  +  l  o 

Ex.  4.  r      dx 


J  a2 


a2  +  b2x2 


Looking  among  the  standard  forms,  it  appears  that  form  9  is  the  one  to 
which  the  integrand  may  be  reduced.  To  get  1  as  the  first  term  of  the  denom- 
inator, we  divide  the  denominator  by  a2,  and  thus  obtain 


1  C      dx        _  1   C        dx 


a    J 

We  observe  that  -  x   is  the  variable,  whence   the   numerator  must  take 
a 

the  fonn  dl -  x  )  =  -  dx.     Therefore  introducing  the  factor  -  in  the  numera- 
\a    )     a  a 

tor  and  the  neutralizing  factor  -  outside  the  integral  sign,  we  have 

b 

a      I  r      a  1  bx  ,    _ 

r  '-sl n — rr,  =  -r  arc  tan \- C. 

b     <*2Ji>(bxy      ab       >       a 

Ex.  5.  f    («-*)<*»    . 

J3x2-6x  +  5 

It  will  be  observed  that  d(S  x2  —  6  x  +  5)  =  fi(x  —  1)  dx  ;  hence,  except  for 
the  factor  6,  the  numerator  of  the  integrand  is  the  differential  of  the  denomi- 
nator.   By  the  introduction  of  this  factor  the  integral  reduces  to  form  13, 

namely,   f —  •     Thus,  we  have 
J  u 

C  (x-\)dx    =lC6(x-l)dx  =  ird^x2-6x+6)  =  ll     ^-2-6 x+5)  +  C. 
J3x2-6:k  +  5     6J3x2-6z  +  5    (iJ     3z2-6a;+5       6    &v 

In  some  cases  the  integral  can  be  reduced  to  the  algebraic  sum 
of  several  standard  integrals. 


Art.  70]  INTEGRATION  BY  INSPECTION  107 


ex.6.  fVl+xdx^f-i±^_dx  =  r_^_  +  r_^_. 

J  Vl  -  x         J  Vl  -  x2         J  Vl  -  x2     J  VI  -  x* 

Now  f — ==  =  arc  sin  as,  (Form  8) 

•J  Vl  -  x2 

and  f    g<to     =-l  C(i_x2)-a(_2a;dx)  =  --  f  (1  -  x2)~i?(l  -  *2) 
J  Vl  —  x'2        % J  %  J 

=  _(l_a;2)^  (Form  1) 

Hence,  f  Vl  +  x  dx  =  arc  gin  ^  _  ^/l  _  x2  +  C. 

J  Vl-x 

If  the  integrand  is  a  rational  fraction  having  the  denominator 
of  the  same  degree  as  the  numerator,  or  of  lower  degree,  perform 
the  division  before  integrating. 


Ex.  7.  C^zl  dx. 

Jx-3 


dx 


We  have  °^zl=x  +  3+  — — 

x-S  x-S 

Hence,  f  ^  -  7  dx  =  (xdx  +  3  f  dx  +  2  f 

=  iz2  +  3a:  +  21og(a;-3)+  C. 

The  following  examples  show  that  it  is  often  possible  to  reduce 
the  given  integrand  to  a  standard  form  by  writing  the  denomina- 
tor as  the  sum  or  difference  of  two  squares. 

Ex.8.      f  g  =f       d^~^       ^arcsin^g-4-C. 

JV7+6x-x2     J  V42  _ (X  -  3)2  4 

Ex  9       r         ^         _  f    ^ (a?  +  2) 

J5-4x-x2     J32-(z  +  2)2 

=  ll0gTzi+C^     if(^  +  2)2<32, 

=  Ilog*±i+C,     if  (x  +  2)2>32. 
6       ic  —  1 


Ex.  10. 


r  dx  C      d(x  +  2) 

•J  Vc2  +  4x  +  8      *  V(x  +  2)2  +  4 


log(x  +  2+Vx2  +  4x  +  8)+  C. 


108  INTEGRATION  [Chap.  VI. 

EXERCISES 

In  the  following  examples  determine  by  inspection  the  proper  standard 
form,  find  the  function  that  replaces  u  in  that  form,  and  perform  the 
integration. 

dx_ 

.  a- 
"cos  6  dd 


1.  ((x  +  aydx.  2.    iJ2x{x2-arfdx.  3.    (J>^ 

a  To     ^ -,  «-     fcos0d0  r 

4.  J2xe*  dx.  5.   j  ^^-'  6.  J  cos20sin0cZ0. 

_  rare  tan  x  dx  g     C       m  dx  n 

'  J      1+*2  "  J  Vl-m2z2'  9*  Js 


sec2  0  tan  0  <Z0. 

10.   Explain  why  the  integration  in  Ex.  9  may  lead  to  either  |sec20  or 
\  tan2  6. 

In  the  following,  reduce  the  integrands  to  standard  forms  by  the  introduc- 
tion of  proper  factors,  and  integrate. 

to     i       x  dx 


u  /(«  +  *>**.  «./^-6.  13.  J 


Va2  +  x2 
16.    \am*x2xdx. 

In  the  following,  reduce  the  integrands  to  standard  forms  by  writing  the 
denominator  as  the  sum  or  the  difference  of  squares,  and  integrate. 

17      f         dx  18      C  dx 

Jz2  +  2x-f5  '    J  V9  +  8x-x2 

rife 


19 


r      dx  20.  f— 

J   >A>2  J.firJ-1ft  ^  11 


Vx2  + 6^  +  10  Jll  +  10x-a2 

Verify  the  following  integrations. 

21.  f      ^       =  arc  sin  -  +  C  =  -  arc  cos  -  +  C". 

J  Va2  -  x2  a  a 

22.  f     ^      =  1  arc  tan  ?  +  C. 

J  a*  +  x'2     a  a 

23.  f — — =  larcsec-+  Q  =  -1arccsc-+  C. 

J  xVx2  -  a2     «  «  a  a 

The  integrals  of  exercises  21-23  may  be  regarded  as  standard 
forms  and  should  be  memorized. 

71.    Integration  by  substitution.    The  reduction  of  a  given  inte- 
grand to  a  standard  form  is  sometimes  most  easily  effected  by  the 


Art.  71]  INTEGRATION   BY   SUBSTITUTION  109 

substitution  of  a  new  variable.     The  following  examples  illustrate 
this  method. 

Ex.  1.  I-?*—. 

J     *  _£ 

e2  +  e    2 

x  dz 

Let  e2  —  Z'  then  dx  =  2  —  j  and  tne  integral  takes  the  form 

'  z 

r 2dz _  2  C_dz_  _  2  arc  tan         ^  _  2  arc  tan  g2  + 

J  2(2  +  2-1)  J  Z2+l 

Ex.  2. 


a 


J 


X2Vl  —  X'2 


Let  x  =  cos  0 ;  then  dx  =  —  sin  0  d0,  V 1  —  x2  =  sin  0,  and  the  integral  takes 
the  form 

J  cos2  0  sin  0         J  x 

The  substitution  x  =  -  may  also  be  used.     The  student  may  work  out  the 
details. 

As  additional  exercises,  the  derivations  of  forms  14  to  19  are 
given. 

Ex.  3.    Since  — — -  =  —  f  — —  ^  ,  we  have,  if  w2  >  a2, 

u2  —  a2     2a\a  —  a     u  +  a) 

r    du    _  i   r_dw_  _  j_  r_^_ 

Jw2  —  a2     2aJ  w-  a     2fJ«-fa 

=  ±[\og(u  -  a)  -  log<>  +  «)]  =~l0^  l±^' 
2a  2  a        u  +  a 

The  student  may  derive  the  second  integral  of  form  14  and  show  that  it 
applies  when  u2  <  a2. 

Ex.  4.   To  integrate    f       du         put  u2  ±  a2  =  z2.    Then  2udu=2zdz, 
-*  Vn2  ±  a2 
whence  du  =  dz  =  du  +  dzm 

z       u        u  +  z 

Therefore,  C       du       =  f  **  =  f  ^LdL^  =  iog(M  +  *) 

^  Vw2  ±  a2      J  z       J    u  +  z 

=  log(t<  +  Vm2  ±  a2) . 

Ex.  5.  ftaix  dx  =  fBin*cte=_  Cdjcosx) 

J  J     cosx  J     cosx 

=  —  log  cos  x  +  G  —  log  sec  x  +  C. 


110  INTEGRATION  [Chap.  VI. 

Ex.  6.  foot x  dx  =  f  <^L^  =  Cdjsinx)  =  {     ^ %  +  Q 

J  J     sin  x        J     sin  x 

Ex.  7.  j  esc  x  dx. 


Let  z  =  tan  -  ;  then  esc  x  =  — — — ,  and  dx  = 


2 '  2z  1  +  z2 

Substituting  these  values,  we  get 

f  csc  x  dx  =  f  —  =  log  z  +  C  =  log  tan  -  +  C. 

Ex.  8.  \  sec  x  dx. 

Making  the  same  substitution  as  in  Ex.  7,  the  integral  reduces  to  the  form 


dz 
—      Hence,  we  have 

22 


1  +  tan  - 


(eecxdx  =  2(^ — -=log-±-?+  C  =  log  -  + 

j  j  i-z  l-z  1  _  tan  ^ 

=  logtan  g  +  j]+  a 


2 


EXERCISES 

1.     T_J^_,  letz  =  a  +  5x  2.    f dx  let  a;  =  a  sec  0. 

Ja  +  bx  Jx2Vx2-a* 

3.     f *5 ,    let*:=asin0.  4.     f     '*»    • 

J(a*-xrf  J  y/x-2 

5.  f ^ f    lets2  =  2sc+l. 

•/XV/2J5+  1 

6.  Verify  the  integrations  given  in  examples  21,  22,  and  23  of  the  preced- 
ing article  by  the  substitutions  x  =  a  sin  0,  x  —  a  tan  0,  and  x  —  a  sec  0, 
respectively. 

C         dx 

7.  Evaluate  the  integral    I  —  by  means  of   the  substitution 

a  =  a(l-cos0).  V2ax-x2 

72.  Character  of  the  integration  process.  There  is  a  funda- 
mental difference  between  differentiation  and  integration.  The 
former  is  a  direct,  the  latter  an  inverse  process.  As  we  have 
seen,  when  once  the  derivatives  of  the  elementary  functions  have 
been  found,  the  derivatives  of  any  functions  expressed  in  terms 


Art.  72]    CHARACTER   OF  THE    INTEGRATION  PROCESS  111 

of  these  functions  can  usually  be  obtained  by  one  or  more  direct 
operations.  On  the  contrary,  it  is  not,  in  general,  possible  to 
determine  the  integral  of  a  function  from  the  known  integrals  of 
the  elementary  functions  in  terms  of  which  the  given  function  is 
expressed.  There  are  no  general  methods  for  expressing  the  in- 
tegral of  a  product  or  a  quotient  of  functions,  or  of  a  function  of 
a  function  in  terms  of  the  integrals  of  the  component  functions. 
Hence  the  process  of  integration  consists,  not  in  a  series  of  direct 
operations  carried  out  in  accordance  with  a  general  method,  but 
rather  in  attempts  to  reduce  the  given  function  to  a  form,  the 
integral  of  which  is  known. 

Integration,  then,  is  largely  a  question  of  judgment  in  attack- 
ing the  problem.  Specific  rules  cannot  be  given,  but  the  follow- 
ing general  directions  will  be  found  useful.  First  inspect  the 
integrand  carefully  and  determine  whether  it  is  a  standard  form. 
Frequently  the  integrand  is  merely  a  disguised  standard  form 
and  needs  only  a  rearrangement  of  its  terms  to  be  recognized  as 
such.  If  the  integrand  is  not  in  a  standard  form,  see  if  it  cannot 
be  made  to  assume  such  a  form  by  the  introduction  of  a  constant 
factor.  Do  not  neglect  the  neutralizing  factor  outside  the  in- 
tegral sign.  If  this  device  does  not  appear  to  be  effective,  try 
the  substitution  of  a  new  variable. 

MISCELLANEOUS  EXERCISES 
Integrate  the  following : 
1.     froz—3<&.  2.     ((3x-x*)~%(S-2x)dx. 


C     x*dx  4     C  (x-S)dx 


(Xs  -  a3) 

f     (*-«)**    .  6.     (xe*> 
J  Vz2  -  6  x  +  1  J 

f  cos2  0  sin  6  dd.  8.     ( 


r(2  +  Bx)dxm  r 


dx. 

cos  6  dd 
sin30 
dx 


±1      C    x*dx  .      \  12      C     2*3-53+!      ^ 

J  3  xs  -  5  J  i ' 


(ax  +  b)% 

2x*-5 

x4-5x2  +  2x-7 


112  INTEGRATION  [Chap.  VI. 

13      C  dx  14      C  x3  dx  t 

J  x*  +  6  x  +  13*  '  •  J  x  +  1 " 


15.     f^f-1^.  16.     f- 

J  xs  +  x  J  a 


17 
19 


sin  x  dx 
x3  +  x  J  a  +  b  cos  as 

f  e  cos  x  sin  j.  tfx.  18.     f     ~  x  dx. 

J  J  1  +  x'2 

f  cfo  t  20      r  (fa 

Jl+3x  +  2z2'  '   J  VU-Sx-x2' 


21.     f    ;      *  22.     f_*L 

^  Vz2  +  6  x  +  1  J  22  - 

23.     M|-.  24.     f_ 

J  4  -  3  6'2  J 


dx 


25 


xVl  —  log2  a; 

.     f         ds        .  26.     f  dM        — 

J  \/2  as  +  s2  ~*  v  (m  +  wi)2  -  n2 


27    r_^j^_,  28    fvi-*^. 

J  Va8  -  x8  ^  Vl  +  x 

29      f  e~x^  30      fO  +  2)<frT 

*   Jl  +  e-*  '     *      xVx 

31.  Given  y  =   (     xdx     ;   knowing  that  y  =  0  when  a;  =  0,  find  the 

■J  Va2  —  a;2 
integral,  determine  the  constant  of  integration,  and  thus  find  the  relation 
between  x  and  y. 

32.  In  the  preceding  example,  find  the  relation  between  x  and  y  if  y  =  0 
when  x  =  a. 

33.  A  curve  passes  through  the  point  (0,  2),  and  the  general  expression 
for  its  slope  is  — Find  the  equation  of  the  curve. 

34.  Find  the  equation  of  a  curve  that  passes  through  the  point  (1,  0)  and 

whose  slope  is  given  by  the  expression  —  +  bx. 

x2 

35.  In  a  certain  type  of  motion  the  relation  —  =  dt  holds,  v0 

\/v02  -  W 
being  the  initial  velocity.     Express  the  distance  s  as  a  function  of  the  time  t, 
knowing  that  s  =  0  when  t  =  0. 


CHAPTER   VII 

SIMPLE   APPLICATIONS    OF   INTEGRATION 

73.  Curves  having  given  properties.  It  has  been  shown  that 
the  value  of  the  derivative  of  a  function  y  =f{x)  for  a  particular 
value  of  the  variable  gives  the  slope  of  the  tangent  to  the  curve 
representing  the  function  for  the  value  of  the  variable.  Further- 
more, it  has  been  shown  that  the  derivative  enters  into  the  equations 
of  the  tangent  and  normal  to  the  curve  and  into  the  expressions 
for  the  length  of  the  tangent,  the  length  of  the  normal,  the  sub- 
tangent,  and  the  subnormal.  We  have  so  far  considered  problems 
in  which  from  a  given  function  the  derivative  was  found,  and 
from  this  derivative  the  desired  property  of  the  curve  was 
obtained.  We  shall  now  consider  inverse  problems  in  which  a 
property  of  the  curve  is  given,  and  from  it  the  equation  of  the 
curve  is  to  be  obtained.  These  problems  naturally  involve  the 
evaluation  of  an  integral. 

Ex.  1.     Determine  the  equation  of  a  curve  at  every  point  of  which  the 
slope  is  equal  numerically  to  one  half  the  abscissa. 
We  have  here  Dxy  =  ±x, 

whence  y  =  \\  x  dx  =  \  x2  +  C. 

This  is  the  equation  of  a  parabola  having  the  F-axis  as  the  axis  of  the  curve. 
By  giving  the  constant  C  different  values,  we  get"  a  series  of  parabolas  having 
the  same  slope  for  the  same  value  of  x. 

Ex.  2.  Find  the  equation  of  the  curve  whose  polar  subtangent  has  a  con- 
stant length  a. 

In  this  case,  we  have  p2  D  6  =  a, 

Therefore,  0=(ad2  =  -Z+C, 

J     p2         p 


C- 
113 


114  SIMPLE   APPLICATIONS   OF   INTEGRATION    [Chap.  VII. 


EXERCISES 

1.  Find  the  equations  of  the  curves  whose  slopes  are  respectively : 

3  ;   I  x2  ;   mx  ; ;    ax  —  b  ; 

x2  ax8 

2.  Find  the  equation  of  the  curve  whose  slope  at  any  point  is  double  the 
abscissa  at  that  point. 

3.  Find  the  equation  of  a  curve  whose  subnormal  is  b  times  the  ordinate. 

4.  Find  the  equation  of  the  curve  whose  subnormal  varies  inversely  as 
the  ordinate. 

5.  Find  the  polar  equation  of  the  curve  for  which  tan  \p  =  kp. 

6.  Find  the  polar  equation  of  the  curve  whose  polar  subnormal  has  a 
constant  length  m. 

7.  Find  the  equation  of  the  curve  which  passes  through  the  point  (2,  3) 
and  whose  tangent  has  the  slope  3  &  +  5. 

8.  Find  the  equation  of  the  curve  whose  slope  at  any  point  is  propor- 
tional to  the  ordinate  at  that  point. 

9.  Find  the  equation  of  the  curve  whose  subtangent  has  the  constant 
value  a. 

10.   Find  the  polar  equation  of  a  curve  for  which  the  angle  \p  is  constant. 

74.    Rectilinear  motion.     The  relations  between  the  time,  speed, 
acceleration,  and  space  traversed  by  a  point  moving  in  a  straight 

lineare:  v  =  Dts,    a  =  Dtv. 

We  have,  therefore,  the  inverse  relations 

v  =  Dt~la=  Cadt,  (1) 

s  =  Drlv=  Cvdt.  (2) 

In  the  direct  problem,  starting  with  the  space  traversed  as  a 
function  of  the  time,  we  were  able,  by  taking  derivatives,  to  find 
the  speed  and  acceleration..  In  the  inverse  problem,  having  the 
acceleration  given  as  a  function  of  the  time,  we  can  by  integra- 
tion determine  first  the  speed,  then  the  space  traversed. 

A  relation  that  is  useful  in  certain  problems  is  the  following : 

c,.  dv        -,  ds 

Since  a  =  —  and  v  =  — , 

dt  dt 


Art.  74]  .    RECTILINEAR   MOTION  115 

we  get  by  eliminating  dt, 

vdv  =  a  ds,  (3) 

whence  v2  =  2  I  ads.  (4) 

Ex.  1.  A  point  moves  in  a  straight  line  with  a  constant  acceleration  a. 
The  speed  and  the  space  passed  over  at  the  end  of  a  given  time-interval  are 
required. 

We  have  v  =  \adt  =  at-\-  C.  (a) 

To  determine  the  constant  C,  let  v0  denote  the  initial  speed,  that  is,  the  speed 
when  t  =  0.     Substituting  these  values  in  (a),  we  have 

v0  =  0  +  C,  or  C  =  v0  ; 

hence,  (a)  becomes  v  =  at  +  v0.  (b) 

We  have  further  s  =  I  v  dt  =  \  (at  +  v0)  <^, 

whence,  performing  the  integration,  we  obtain 

s  =  $aP  +  vot+  C".  (c) 

To  determine  the  constant  C",  let  s0  denote  the  initial  space,  that  is,  the  space 
when  t  —  0.     Substituting  in  (c) ,  we  have 

s0  =  0  +  C",  or  C  =  So, 

and  we  have  finally  s  =  ^at2+  v0t  +  s0.  .  (d) 

An  important  application  of  these  laws  is  to  the  case  of  freely 
falling  bodies.  By  observation,  it  is  found  that  a  body  falling 
freely  in  a  vacuum  at  a  given  point  on  the  earth's  surface  has  a 
constant  acceleration.  This  acceleration,  while  constant  for  any 
one  place,  varies  for  different  localties  within  small  limits.  Its 
value  may  be  taken  as  32.2  ft./sec.2. 

If  the  body  falls  from  rest,  we  have  v0  =  0  and  s0  =  0.  Denoting 
by  g  the  constant  acceleration,  we  have,  therefore, 

v  =  gt,  (5) 

s  =  igf.  (6) 

Eliminating  t  between  (5)  and  (6),  we  get  as  a  third  relation, 

v2  =  2  gs, 
or  v  =  V2gs,  (7) 

which  might  also  be  obtained  from  (4). 


116  SIMPLE   APPLICATIONS   OF   INTEGRATION    [Chap.  VII. 

Ex.  2.  Investigate  the  motion  of  a  body  projected  vertically  upward  from 
the  earth's  surface  with  an  initial  velocity  of  b  feet  per  second. 

We  have  here  v0  =  b  and  s0  =  0.  The  acceleration  is  g,  but  in  this  case  is 
opposite  to  the  direction  of  motion  ;  hence  a=—  g.  From  the  fundamental 
relations  (1)  and  (2),  we  have  upon  integration 

v  =-  gt  +  b, 

The  body  will  reach  its  highest  point  when  v  =  0,  that  is,  when 
0  =—  gt  +  ft,  or  t  =  -i 

and  the  height  h  to  which  it  rises  is  found  by  substituting  this  value  of  t  in 
the  second  equation.     Thus 


h=—ig[-\   +o--  =  ^— 

*yW/         g    %g 


75.   Rotation  about  a  fixed  axis.     Corresponding  to  the  direct 

relations 

o)  =  Dt0,         a  =  Dtu>, 

we  have  the  inverse  relations 

(o  =  Dt-1a=   fa  (it,  (1) 

0  =  Dt-'«>=  Ciodt,  (2) 

and  starting  with  the  angular  acceleration  a  as  a  given  function 
of  the  time,  we  can  by  integration  determine  the  angular  speed 
and  the  angle  swept  through  by  a  given  radius.  For  the  case  in 
which  the  angular  acceleration  is  a  constant,  say  aQ,  we  readily 
derive  the  following  formulas  for  <o  and  0 : 

<»>  =  (V  +  wo-  (3) 

6  =  \a4'i  +  ^t  +  0{,  (4) 
As  in  Art.  74,  we  have  also  the  useful  relation 

a>d<o  =  add.  (5) 

7C>.  Motion  of  a  projectile.  A  body  is  thrown  with  an  initial 
velocity  v0  at  an  angle  a  with  the  horizontal.  Neglecting  the 
resistance  of  the  atmosphere,  the  path  of  the  body  is  to  be 
determined. 


Art.  70] 


MOTION   OF  A  PROJECTILE 


117 


Let  the  plane  containing  the  path  be  taken  as  the  XF-plane, 
with  the  F-axis  vertical.  Then  the  X-  and  F-coinponents  of  the 
initial   velocity   v0  are   respectively   vQ  cos  «   and   v0  sin  a.     The 


Fig.  25. 


X-component  remains  constant  throughout  the  motion,  that  is, 
the  X-component  ax  of  the  acceleration  of  the  body  is  zero.  The 
F-component  ay  of  the  acceleration  is  —  g,  as  in  the  case  of  a 
body  projected  vertically  upward.     We  have  therefore 

dvx 
dt 


a* 


0, 


_dv1 


a«=a=-g> 


■y,=  Cj  =  r0oos  a. 


(1) 
(2) 


To  determine  02,  we  have  at  the  beginning  of  the  motion 
vy  =  v0  sin  a,  when  t  =  0  ;  hence  C2  =  v0  sin  a,  and  vy  =  v0  sin  a  —  gt. 
To  find  the  X-  and  F-coordinates  of  the  position  of  the  body  at 
the  time  t,  we  have  ^ 


v,  =  —  =  v0  cos  a, 
dt 


x  =  v0t  cos  a ; 
y     eft 


0*j 


whence 

and  »f  =  -*■  =  i?0  sin  « 

whence  y  =  vj,  sin  a  —  -J-  gtf2. 

The  constants  of  integration  are  readily  found  to  be  zero. 
Eliminating  t  between  (3)  and  (4),  the  resulting  equation, 

y  =  x  tan  a  -  — -2 — --, 
2  V  cos^  a 

is  the  equation  of  the  path. 


(3) 


(4) 


(5) 


118  SIMPLE   APPLICATIONS   OF   INTEGRATION    [Chap.  VII. 


EXERCISES 

1.  Find  an  expression  for  the  velocity  v  and  distance  s  when  the  accelera- 
tion is  given  by  the  relation : 

(a)  a  =  m-nt\  (6)  a  =- nttf  cos  kt.  (c)  a  =^2  (e«-  e~«). 

2.  If  the  speed  of  a  point  moving  in  a  straight  line  is  given  by  the  relation 
v  =  10 1  —  t2,  find  expressions  for  the  acceleration  and  the  distance  traversed, 
starting  from  rest,  at  the  end  of  t\  seconds. 

3.  In  Ex.  2  find  the  time  that  will  elapse  before  the  point  comes  to  rest, 
the  distance  the  point  has  moved  in  that  time,  and  the  final  acceleration. 

4.  A  hoisting  drum  has  an  angular  speed  w0  =  24  rad./sec.  A  brake  is 
applied  in  such  a  way  as  to  produce  a  retardation  (negative  acceleration)  of 
3  rad./sec.  Find  the  time  required  to  bring  the  drum  to  rest  and  the  number 
of  revolutions  made  after  the  brake  is  applied. 

5.  Show  that  the  range  of  a  projectile  on  a  horizontal  plane  is 

B=^sm2a. 
9 

6.  Show  that  the  velocity  of  a  projectile  at  any  instant  is  v  =  vW  —  2  gy. 

7.  A  stone  is  thrown  horizontally  with  an  initial  velocity  Vq  from  the  top 
of  a  cliff.     Find  the  path  followed. 

8.  If  the  top  of  the  cliff  is  100  feet  above  the  level  of  a  stream  150  feet 
wide  at  its  base,  what  initial  velocity  is  required  to  carry  the  stone  across  the 
stream  ? 

77.  Harmonic  motion.  According  to  the  laws  of  mechanics, 
the  acceleration  of  a  body  of  given  mass  is  proportional  to  the 
force  acting  on  it.  Hence,  since  it  is  usually  the  force  that  is 
known,  the  character  of  the  motion  is  specified  by  the  law  ac- 
cording to  which  the  acceleration  changes ;  and  the  velocity  and 
distance  traversed  are  obtained  by  one  and  two  integrations, 
respectively.  The  acceleration  a  may  be  given  as  a  function  of  t, 
of  v,  or  of  s.  In  this  article  and  those  immediately  following 
we  shall  consider  a  few  important  examples  of  motion  following 
various  laws. 

When  a  point  moves  in  such  a  way  that  the  negative  accelera- 
tion is  proportional  to  the  distance  of  the  point  from  a  fixed  origin 
0,  the  motion  is  said  to  be  simple  harmonic.     In  this  case,  we  have 

a  =  -k2s,  (1) 


Arts.  77,  78]       MOTION  IN   A   KESISTING   MEDIUM  119 

whence  from  the  relation  vdv  =  a  ds,  [Art.  74,  Eq.  (3)] 

vdv  =  —  k2s  ds. 
Integrating,  we  obtain 

v*=C-fcV. 

To  determine  the  constant  C,  let  v0  denote  the  velocity  at  the 
origin    0;    then   v  =  v0   when    s  =  0,    and    consequently    C  =  V- 

Hence  v2  =  v<?  -  tfs2.  (2) 

Evidently  v  =  0  when  v2  —  tfs2,  that  is,  when  s  =  ±  ^  •     It  follows 

k 

that  the  point  oscillates  between  the  points  —  and  —  —  equidis- 
tant from  the  origin. 

The  relation  between  the  distance  traversed  and  the  time  is 
obtained  from  the  relation 

ds 

whence  — ==z=  =  dt. 

vV-W 

Integrating,  we  have 

^arcsin  —  =  t  +  C.  (3) 

K  V0 

If  we  make  t  =  0  when  the  point  is  at  the  origin  and  s  =  0,  then 
(7=0,  and  we  have 

s=-°sinto.  (4) 

k 

Erom  (4),  as  from  (2),  we  can  see  the  vibratory  character  of  the 

q 

motion;  for  as  kt  takes  successively  the  values  0,  ^,  -n-,  —,  2irf 
etc.,  sin  kt  takes  the  values  0, 1,  0,  —  1,  0,  etc.,  and  s  the  values  0, 

|«,  0,-^,0,  etc. 

k  k 

78.  Motion  in  a  resisting  medium.  When  a  body  moves  in  a 
medium  as  air  or  water,  it  encounters  a  resistance  that  is  de- 
pendent upon  the  velocity  v.  For  small  values  of  v,  this  resistance 
is  approximately  proportional  to  the  first  power  of  v,  while  for 
higher  velocities  it  is  more  nearly  proportional  to  v2. 


120  SIMPLE  APPLICATIONS   OF   INTEGRATION    [Chap.  VII. 

1.    Let  a  body  having  an  initial  velocity  v0  be  subjected  to  a 

resistance  proportional  to  v ;  then  the  acceleration  is  —  kv,  where 

k  is  a  constant,  and  we  have 

dv         7  , . . 

a  =  jr-kv,  (1) 

whence  dt  = , 

kv 

and  t  =  hog--\-Cv  (2) 

k        v 

To  determine  G1}  we  have  v  =  v0  when  £  =  0.    Therefore  C1  =  -  log  v0, 
and  (2)  becomes 


t  = 

AS           V 

(3) 

whence 

v  = 

:  v0e~*\ 

(4) 

To  find  i 

a  relation  between  s 

and  £,  we 

have  from 

(4) 

ds  _ 
dt" 

:  v0e-to, 

whence 

s  = 

+  c2. 

(5) 

Putting 

s  =  0,  when  t  -. 

=  0,  we 

j  find  C2  = 

=  ~2,  whence  (5) 
ft 

becomes 

s  = 

3(1 -e" 

-")■ 

2.  A  body  falling  from  a  height  to  the  earth  encounters  a 
resistance  from  the  atmosphere  proportional  approximately  to  the 
square  of  the  velocity.  The  acceleration  without  resistance  being 
g,  the  acceleration  when  the  resistance  is  taken  into  account  is 
evidently  g  —  k2v2.     We  have  then 

also  from  the  relation  vdv  =  a  ds, 

vdv  =  (g-k2v2)ds.  (7) 

From  (6),    t  =  f -*      =  J- log  ^+^  +  %  (8) 

Jg-k2v2     2V^       Vg-kv 


Art.  78]  MOTION  IN  A   RESISTING  MEDIUM  121 

and  from  (7)  .-  J_^_  =  -ilog(«,_*V)+  C,  (9) 

If  we  take  the  initial  velocity  as  0  when  s  =  0  and  t  =  0,  we  find 
the  constants  to  be  Cx  =  0,  (72  =  - — -  log  g.     Hence  from  (8),  we  get 

fc  Veav"  +  iy 
and  from  (9)  ..^.fcg^,  (11) 


whence 


^  =  £(l_.e-a»).  (12) 


As  s  increases  indefinitely,  the  velocity  v  approaches  the  limit- 

ing  value  — -£.     Thus  a  body  falling  from  a  great  height,  as  a 

raindrop,  approaches  the  earth  with  practically  this  limiting 
velocity.  Observe  that  the  same  result  is  obtained  from  (10) 
when  t  is  increased  indefinitely. 


EXERCISES 

1.  Show  that  the  time  of  a  complete   oscillation   in   harmonic  motion 

is**. 

k 

2.  If  a  point  moves  in  a  circle  with  constant  speed,  the  projection  of  the 
point  on  a  diameter  moves  on  the  diameter  according  to  the  law  of  simple 
harmonic  motion.  Prove  this  statement,  and  show  that  the  constant  k  in 
the  equation  of  Art.  77  is  the  angular  speed  of  the  radius  that  joins  the 
moving  point  to  the  center. 

3.  A  man  is  rowing  in  still  water  at  a  speed  of  m  ft.  per  second.  If  he  sud- 
denly stops  rowing,  find  the  law  according  to  which  the  boat  continues  to 
move,  assuming  that  the  resistance  of  the  water  is  proportional  to  the  speed. 

4.  In  Ex.  3  find  the  time  that  must  elapse  before  the  boat  comes  to  rest. 
Give  a  reason  for  the  absurdity  of  the  result. 

5.  Show  that  the  speed  in  harmonic  motion  is  expressed  as  a  function  of 
the  time  by  the  equation  v  =  vq  cos  kt. 


122  SIMPLE  APPLICATIONS   OF  INTEGRATION    [Chap.  VII. 

6.  If  the  retarding  effect  of  fluid  friction  on  a  rotating  disk  is  propor- 
tional to  the  angular  speed  «,  that  is,  if  —  =  —  &w,  show  that  <a  =  w0e-^, 
where  a>o  is  the  initial  angular  speed. 

7.  Find  an  expression  for  <a  when  the  retardation  is  proportional  to  the 
square  of  the  angular  speed  w,  which  is  approximately  true  for  a  very 
rapidly  revolving  disk,  as  a  gyroscope. 

79.  Physical  problems  involving  exponential  functions.  If  a 
function  has  the  general  form 

V  =  eax, 

then  Dxy  =  aeax  =  ay ; 

that  is,  the  rate  of  change  of  the  function  is  proportional  to  the 
function  itself.  Many  natural  phenomena  follow  this  law,  which 
has  been  called  by  Lord  Kelvin  the  compound  interest  law.  One 
example  of  this  law  has  been  shown  in  the  motion  of  a  body  in 
a  resisting  medium.     The  following  are  other  illustrations  : 

(a)  Atmospheric  pressure.  It  is  well  known  that  the  pressure  of  the 
atmosphere  decreases  as  the  distance  from  the  earth's  surface  increases. 
Assuming  the  temperature  of  the  atmosphere  to  be  constant,  the  rate  of 
change  of  pressure  with  the  height  at  any  given  height  is  proportional  to  the 
pressure  at  that  height ;  that  is, 

d?  =  -kp. 
dh 

The  negative  sign  is  used  because  p  decreases  as  h  increases.     We  have  then 

<lP  =  -kdh, 
P 

and  integrating,  the  result  is 

\ogp  =  -kh  +  C. 

To  determine  the  constant  of  integration  C,  let  po  denote  the  pressure  at  the 
earth's  surface  where  h  =  0.     Then 

logpo  =  0  +  C,  or  C  =  logpo- 
Substituting  this  value  of  C  and  transposing,  we  have 

kh  =  \ogp0-  log/)  =  log^, 
P 
whence  ^  =  e*» 

P 

or  p=p0e'kh. 


Art.  79]  EXPONENTIAL   FUNCTIONS  123 

This  formula  gives  the  pressure  at  any  definite  height,  when  the  constant  k 
is  known. 

(b)  Newton's  law  of  cooling.  A  body  has  a  temperature  r  which  is 
higher  than  the  temperature  t0  of  the  surrounding  medium.  The  rate  at 
which  it  cools  is  approximately  proportional  to  the  difference  in  temperature 
t  —  r0  ;  that  is,  approximately, 

dt 
where  It  is  some  constant.     We  have  then 
dr 

T  —  T0 

whence,  by  integration, 

log  (t  -  r0)  =  -  kt  +  C. 
Let  Tj  denote  the  temperature  when  t  —  0  ;  then 

C^logC^-To), 

and  substituting  this  value  of  C,  we  get 


or 


kdt, 


T    -T0 

Hence,  we  have  t  =  t0  +  (r.  —  t0)  e~**, 

from  which  t  can  be  found  for  any  time  interval. 

(c)  diversion  of  sugar.  Cane  sugar  in  solution  is  decomposed  into  other 
substances  through  the  presence  of  acids.  The  rate  at  which  the  process 
takes  place  is  proportional  to  the  mass  of  sugar  still  unchanged.  Thus,  if  s 
is  the  original  mass  of  sugar  and  x  is  the  mass  inverted,  the  rate  of  inversion 
at  this  stage  of  the  process  is  proportional  to  the  unchanged  mass  s  —  x.  The 
equation  expressing  this  law  is 

%  =  k(s-x). 
at 
Integrating,  we  get 

—  log  (s  —  x)  =  kt  +  G. 

To  determine  the  constant  of  integration  O,  we  make  use  of  the  fact  that 
x  =  0  when  t  =  0,  whence  C  =  —  log  s.  Substituting  this  value  of  C  in  the 
original  equation,  we  have 

log  — —  =  kt, 
s  —  x 


or 


Jet 


S—  X 

Solving  for  jc,  we  get 

x  =  s  (1  —  e-**). 


124               SIMPLE   APPLICATIONS   OF  INTEGRATION   [Chap.  VII. 

MISCELLANEOUS    EXERCISES 
Integrate  the  following. 

1      C  xdx                   2      C  ^dx  3     C   e2xdx 

J  x4  -f-  3*                      Jx2-1  '  JveMH* 

4      C     dx                   5      f       %dx  6     C dx 

J  ex  +  e-x                     J  x>/4a;'2-7  '  ^  (x-a 


)V(x-a)2-b2 

7.  Determine  curves  whose  slopes  are  respectively 

(a)  3x  +  2;     (6)  x^  +  c  ;     (c)  cos  rax;     (<f)  ecx. 

Assume  a  point  on  each  curve  and  thus  determine  the  constant  of  integration 
in  each  case. 

8.  Find  the  equation  of  a  curve  whose  tangent  has  the  slope  4  —  x2  and 
which  passes  through  the  point  (3,  —  2). 

9.  Since  the  specific  heat  c  is  given  by  the  derivative  — " ,  it  follows  that 

dr 

Q  =  (cdr.  (See  Art.  19.)  From  the  experimental  law,  c  =  a  +  br  +  ct2, 
derive  an  expression  for  Q. 

10.  The  equation  E  =  Bi  +  L  —  expresses  for  a  constant  electromotive 

(Mi 

force  E  the  relation  between  the  time  t  and  the  current  i  in  a  circuit  in  which 
an  electric  current  is  flowing.  B  denotes  the  resistance  and  L  the  inductance 
of  the  circuit,  and  both  are  constants.     If  i  =  0  when  t  =  0,  show  that  the 

Rt. 

■pi  i 

value  of  i  at  the  time  £i  is  —  (1  —  e    L). 
B 

11.  If,  in  Ex.  10,  to  is  the  value  of  the  current  at  a  given  instant,  and  the 
electromotive  force  E  is  removed  at  this  instant,  show  that  at  the  end  of  t\ 

JK, 

seconds  after  the  removal  the  value  of  the  current  is  i\  =  ioe    L  . 
Suggestion.     Make  E  =  0,  and  take  t  =  0  when  i  =  i0. 

12.  If  the  acceleration  of  a  point  moving  in  a  straight  line  is  given  by 
the  relation  a  =  k2s,  show  that  the  relation  between  the  distance  and  the 

time  is  s  =  —  (ekt  -  e~kt),  where  v0  denotes  the  initial  speed. 
Z  tc 

13.  A  curve  is  described  by  a  point  moving  along  a  radius  vector  while 
the  radius  is  rotating  about  a  pole.  If  the  velocity  along  the  radius  is  n 
times  the  velocity  at  right  angles  to  it,  what  is  the  polar  equation  of  the 
curve  ? 


Art.  79]  MISCELLANEOUS  EXERCISES  125 

14.  In  the  theory  of  the  bending  of  beams,  x-anti-derivatives  of  the  fol- 
lowing functions  are  required.  Find  these  anti-derivatives  and  determine 
the  constants  of  integration  from  the  conditions  given. 

(a)  f(x)  =  i  w(L  -  x)2,  #_1/0)  =  °.  wnen  x  =  °- 

(6)  /(x)  =  LH  -Lx*+\  x3,  D-lf(x)  =  0,  when  x  =  0. 

(c)  /(x)  =  I  wL2x  -  £  wxs,  D-lf(x)  =  0,  when  x  =  0. 

(d)  /(»)  =  |  w>Zx  -  £  wx2,  D^f(x)  =  0,  when  x  =  L. 


CHAPTER   VIII 
SUCCESSIVE   DIFFERENTIATION   AND    INTEGRATION 

80.  Definition  of  the  nth  derivative.  The  derivative  of  a  func- 
tion is,  in  general,  a  function  of  the  same  independent  variable 
as  the  original  function,  and  it  may  itself  have  a  derivative  if  it 
fulfills  the  necessary  conditions.  The  differentiation  of  the  origi- 
nal function  produces  the  first  derivative ;  the  differentiation  of 
this  first  derivative  gives  the  second  derivative;  the  result  of 
differentiating  this  second  derivative  is  the  third  derivative ;  and 
so  on.  If  the  process  is  repeated  n  times,  the  final  result  is  the 
nth  derivative  of  the  function  with  respect  to  its  independent 
variable. 

As  an  example,  consider  the  function 

/(a?)=a^-2aj  +  5. 

The  first  derivative  is  3  a;2  —  2, 

which  is  also  a  function  of  x.  Differentiating  this,  the  second 
derivative  is  6  x,  which  is  again  a  function  of  x.  The  third  deriv- 
ative is  6,  a  constant,  and  the  fourth  and  each  successive  deriv- 
ative is  zero. 

If  y=f(x)  is  the  original  function,  the  successive  derivatives 
are  also  denoted  by  the  symbols : 

Dxy,  D*y,  Dx*y,  -,  Dxny, 

or  by  %  ^,  ^,  •-.,  ^- 

dx    dx2'  dx3'       '  dxn 

The  symbols  —J,  — ^,  etc.,  are  the  ones  most  frequently  used. 

The  form  of  these  symbols  may  be  explained  as  follows:    Since 

126 


Arts.  80,  81]  IMPLICIT  FUNCTIONS  127 

d  f(x)  d 

J  v  >  =f'(x),  we  may  regard  —   like  Dx  as  an  operator  which, 
dx  dx 

when  applied  to  f(x),  produces  the  derivative  f'(x).     If 

2/=/0»), 

we  have,  therefore,  —  y  =  -^  =  /"'(#), 

'  '  dxy      dx    J  v  J' 

dx\dx)    J    v  h 

dra_r^\i     ,„ ()  ^ 

dx\_dx\dxj]     J     v  " 
For  the  sake  of  convenience  in  writing 

£-(&)  is  abbreviated  to  %  ±[£-(&)~\  to  %  etc. 
dx\dxj  dx~    dx[_dx\dxjj        dx6 

A  general  formula  for  the  nth  derivative  may  be  found  for  cer- 
tain functions  as  shown  by  the  following  examples. 

Ex.  1.     Let  y  =  x?,  where  p  is  a  positive  integer. 
Then  Dxy  =pxP~1, 

Dx2y=p(p-  1)3*-*, 

Dxsy=p(p-l)(p-2)xps, 


Dxny=p(p-l)(p  -  2)  ...  (p-  n  +  1)xp-». 

For  n  =p,  xP~n  =  1  and  Dxp(xp)=p  !;  hence  for  n>p,  the  derivative  is 
zero. 

Ex.  2.    Let  y  =  eax. 

We  have  Dxy  =  aeax, 

Bx2y  =  a2eax, 

Dx*y  =  aseax, 

Dj*y  =  aneox. 

81.  Successive  differentiation  of  implicit  functions.  The  follow- 
ing example  illustrates  the  method  of  procedure  in  finding  suc- 
cessive derivatives  of  an  implicit  function. 


128  DIFFERENTIATION  AND  INTEGRATION    [Chap.  VIII. 

Ex.     Given  f(x,  y)  =  xhj  +  5  y  -  3  X  =  0  ;  find  £-2. 

dx2 


Differentiating  with  respect  to  as,  we  obtain 

2^  +  ^  +  5^-3  =  0, 
dx        dx 


(1) 


whence  dy  =  S-2xy,  () 

dx       z*  +  5  w 

A  second  differentiation  with  respect  to  x  gives 

da;2  (x2  +  5)2 

-  (x2  +  5)(2?/  +  2a;^)-  (3-2a;y)2a; 


(3) 

Substituting  in  (3)  the  expression  for  -^  given !  in   (2) ,  we  obtain  after 
reduction  x 


(x2  +  5)2 

>n  for  ^ 
da; 

d2y  _  6  x2y  -  12  a;  -  10  y 


(4) 
dx2  (a;2  +  5)2  v  J 

82.  Geometrical  and  physical  interpretations  of  the  second  deriva- 
tive.    As  has  been  shown,  the  first  derivative  Dxy  (or  -^  J  gives 

\      dx) 

the  rate  of  change  of  y  with  respect  to  x,  and  expresses  geometri- 
cally the  slope  of  the  curve  which  is  the  graph  of  y  =f(x).  Evi- 
dently,  the    second    derivative   Dx2y  (or  —£ )  gives   the   rate   of 

change  of  the  slope  compared  with  the  rate  of  change  of  the  ab- 
scissa x.  Take,  for  example,  the  function  y  =  mx  -J-  b,  which  is 
represented  by  a  straight  line.  The  slope  is  constant,  and  there- 
fore its  rate  of  change  is  zero.  This  is  shown  by  the  derivatives ; 
for  D^j  =  m  and  D2y  =  0. 

In  the  case  of  a  moving  point,  the  speed  v  is  the  derivative  Dts, 
while  the  tangential  acceleration  a  is  Dtv.  If  now  Dts  be  substi- 
tuted for  v,  we  have 

a  =  Dt(Dts)  =  D2s, 

or  a  =  —  -,  (1) 

dt2'  V  J 

that  is,  the  tangential  acceleration  is  the  second  time-derivative  of 
the  sjmce  s. 


Art.  82]  THE    SECOND   DERIVATIVE  129 

In  the  case  of  rotation  about  a  fixed  axis,  the  angular  speed  is 

dB 
Dt6  =  —  =  w,  and  the  angular  acceleration  is 

(XL 

a=Dta>  =  Dt  (Dt0)  =  D:-6  =  |?;  (2) 

that  is,  the  angular  acceleration  is  the  second  time-derivative  of  the 
angle  swept  over. 

EXERCISES 

Find  the  second  derivatives  of  the  following  functions. 
1.    y  =  a*  +  4a;  — 7.  2.    y  =  xx.  3.    y  =  e^cosx. 


4.    y  =  arc  cos  x.  5.    y  =  Va2  —  x': .  6.   y  =  log(a2  +  x2). 

Find  — -^  for  the  following. 
dx:i 

7.    y  =  xex.  8.    y  =  log(x  —  3).  9.    y  =  x  sin  x. 

10.  If  y  =  sin  x  +  cos  x,  show  that  y  =  £l  =  ^  • 

dx4      dx8 

Find  the  nth  derivatives  of 

11.  y  =  a*.  12.  y  =  -  •  13.  y  =  log  x. 

x 

14.  Find  — ^  from  (a)  the  equation  of  the  parabola  y2  =  'xpx;  (6)  the 

dx2 

equation  of  the  ellipse  a2y2  +  b2x2  =  a2b2. 

Find  — e-  for  the  following  implicit  functions. 
dx2 

15.  x8y  +  3  x2y2  +  xys  =  0.  16.   y2  -  2axy  +  x2  -  c  =  0. 

17.  .rev  —  c  =  0. 

18.  Find  —  from  r 


dff2  1  —  cos  d 

rfir 

19.    Find  —  from  r  =  log  sin  0. 


20.  If  jw"  =  C,  show  that  ^  =  n(n  +  1)  -?• 

dv2  v2 

21.  If  w  =  e«  sin  a;,  show  that  ^  -  2dy  +  2  y  =0. 

dx2        dx 

22.  If  y  =  log  [x  +  VaM^2],  show  that  f|  +  -r*—^  =  °- 

dx2      a2  +  x2  dx 


130  DIFFERENTIATION  AND  INTEGRATION    [Chap.  VIII 

83.  Successive  integration.  As  the  inverse  of  successive  differ- 
entiation, we  have  the  process  of  successive  integration.  Starting 
with  a  function  y  =f(x),  considered  as  an  nth  derived  function, 
a  single  integration  gives  a  new  function,  the  integral.  The  in- 
tegration of  this  second  function  gives  a  second  integral,  and  so 
.on.  The  result  of  n  integrations  is  the  nth  integral  of  the  given 
function. 

For  example,  let  f(x)  =  5  x2. 

Then  |6  x2dx  =  f  x8  +  Ci  is  the  first  integral, 

I  (t  £2  +  Ci)dx  =  T\  x4  +  C\X  +  (72  is  the  second  integral, 

)  (A  **  +  CtoB  4-  C2)dx  =  A  ^5  +  i  Cix2  +  C^  +  Gs  is  the  third  integral, 
etc. 

It  will  be  observed  that  an  integral  contains  the  number  of 
arbitrary  constants  indicated  by  its  order;  thus  the  third  inte- 
gral has  three,  the  fourth,  four,  and  so  on. 

Let  the  successive  integrals  of  f(x)  be  denoted  by  fi(x), 
f2(x),  ••-,/„(#),  respectively;  we  then  have 

/2(aO=J/iO»)d«; 
fs(x)  =  JMX)  dx> etc- 

Ex.  1.     Find  the  successive  integrals  of  y  =  eax. 
/i(x)  =  f/(x)da;  =  (eaxdx=-eax  +  d, 

/,(«)  =  J  (Uu  +  CiW  =i«-  +  Cix  +  C2, 

/a(x)=  f  f \e»  +  dx+  C2W  =  1*»  +  J  daS"+  C2x  +  C3, 

/„(»)  =  —  ettX  +  l-ix'1-1  +  A:2x«-2  +   ...   +*„_!*  +  *„, 

in  which  &i,  &2,  ••-,  kn  involve  the  constants  of  integration. 

Ex.  2.  A  body  falls  with  a  constant  acceleration  g  =  32.2  ft. /sec.2.  If  it 
starts  from  rest,  through  what  distance  will  it  fall  in  10  seconds  ? 


Art.  83]  SUCCESSIVE   INTEGRATION  131 

We  have  here  —^  =  32.2,  (1) 

**  =  (<**  dt  =  f  32.2  dt  =  32.2  t  +  Ci,  (2) 

dt      J  dt1         J 

s  =  f  J|  dt  =  (  (32.2  t  +  Ci)  d*  =  16.1 t2+  Gxt  +  C2.  (3) 

We  may  now  determine  the  constants  Ci,   C2,  from  the  initial  conditions 
of  the  problem.     Since  the  body  falls  from  rest,  we  have  for  t  =  0, 

s  =  0,    ^  =  0. 
dt 

Hence  for  t  =  0,  we  have  from  (2)  and  (3) 

&  =  0,    <72  =  0. 

Therefore,  for  the  time  t,  the  space  passed  over  is  given  by 

s  =  16.1  t2. 

For  t  =  10  seconds,  s  =  1610  feet. 

EXERCISES 

Find  four  successive  integrals  of  the  following  functions. 

1.   y=sinax.  2.    y=x2  —  1.  3.    y=  —  • 


E/L  2  J' 


4.  Find  a  curve  that  passes  through  the  points  (0,  0)  and  (w,  n)  and 

for  which  the  second  derivative  — B   at  any  point  is  A:  times  the  abscissa  at 

dx2 

that  point. 

5.  In  the  theory  of  flexure  of  beams  the  following  equation  occurs : 

d2y 

dx2     EI\ 

E,  7,  M,  i?,.and  w  being  constants.  Derive  an  expression  for  y  and  deter- 
mine the  constants  of  integration  C\  and  C%  from  the  conditions  y  =  0, 
when   x  =  0,  and  y  =  0  when  x  =  I. 

6.  A  point  has  an  acceleration  expressed  by  the  equation  a  =  —  ru>2  cos  ut, 

where  r  and  o>  are  constants.      Derive  expressions  for  the  velocity  and  the 

distance  traveled. 

d2s 
We  have  —  =  —  rw2  cos  <at. 

dt2 

Hence,  v  =  —  =  f  —  dt=-  ru2  f  cos  ut  dt 

dt     J  dt2  J 

=  —  ru  sin  ut  +  C", 


132  DIFFERENTIATION  AND   INTEGRATION    [Chap.  VIII. 

and  s  =  f  —  dt  =  -r(a  ( sin  utdt  +  (  C'dt 

=  r  cos  (at  +  C't  4-  6y", 
which  is  the  law  of  simple  harmonic  motion. 

7.   In  Ex.  6,  determine  the  constants  of  integration  from  the  following 
initial  conditions :  s  =  r  and  v  =  0,  when  t  =  0. 


84.  Maxima  and  minima.  In  Chapter  III  was  discussed  the 
relation  of  the  derivative  to  increasing  and  decreasing  functions 
and  to  turning  points  of  a  curve.  At  a  turning  point,  the  function 
has  the  largest  or  the  smallest  value  of  all  values  in  that  neighbor- 
hood ;  and  we  say  that  it  has  a  maximum  or  a  minimum  value  at 
such  a  point.  A  maximum  value  of  a  function  may  therefore  be 
defined  as  follows: 

A  function  f(x)  is  said  to  have  a  maximum  for  x  =  a,  iff(a)  is 
greater  than  the  values  of  f(x)  for  x  just  preceding  and  just  follow- 
ing x  =  a.     Expressed   symbolically,  f(x)  has   a   maximum   for 

X  =  a'it  /(«)>/(«  ±  K), 

where  h  is  positive  and  takes  all  values  in  the  immediate  neighbor- 
hood of  zero. 

Likewise,  a  function  f(x)  is  said  to  have  a  minimum  for  x  =  a, 
if  f(a)  is  less  than  the  values  of  f(x)  for  x  just  preceding  and  just 
following  x  =  a  ;  in  other  words,  if 

f(a)<f(a.±h), 

for  all  values  of  h  in  the  neighborhood  of  zero. 

A  function  may  have  several  maxima  and  several  minima  in 
any  given  interval.  We  shall  limit  the  present  discussion  to 
functions  having  only  a  finite  number  of  maxima  and  minima 
in  a  finite  interval. 

The  analytic  conditions  for  a  maximum  or  a  minimum  are 
easily  deduced  when  the  function  has  a  derivative.  If  f(a)  is  a 
maximum, /(a?)  must  increase  with  x  just  before  x  =  a  and  de- 
crease as  x  increases  just  beyond.  As  shown  in  Art.  34,  the 
derived  function /'(a?)  must  then  be  positive  for  values  of  x  just 
preceding  x  =  a  and  negative  for  all  values  just  following.  Simi- 
larly, if /(a)  is  a  minimum,  fl(x)  must  be  negative  for  values  of  x 


Art.  84]  MAXIMA  AND   MINIMA  133 

just  preceding  x  =  a  and  positive  for  values  just  following.     We 
may  combine  these  statements  in  the  following  theorem : 

Theorem  I.  f(a)  is  a  maximum  or  a  minimum  of  the  function 
f(%),  according  as  fix)  changes  from  positive  to  negative  or  from 
negative  to  positive  values  as  x,  for  increasing  values,  passes  through 
the  value  x=  a.  Iff'(x)  does  not  change  sign  as  x  passes  through 
x  =  a,  then  f(x)  has  neither  a  maximum  nor  a  minimum  at  x  =  a. 

To  apply  this  test,  substitute  (a  —  h)  and  (a  +  h)  for  x  in  f'(x). 
In  case /(a)  is  a  maximum,/ '(a  —  h)  remains  positive  and/'(a  -+-  h) 
negative  however  small  the  value  of  h  is  taken ;  if  /(a)  is  a  mini- 
mum, then  f'(a  —  h)  remains  negative  and  f'(a  +  h)  positive  for 
arbitrarily  small  values  of  h. 

If  the  maximum  or  minimum  occurs  at  an  ordinary  point  of  the 
curve,  the  tangent  at  the  point  is  parallel  to  the  X-axis,  and  con- 
sequently the  first  derivative  for  this  value  of  x  must  vanish; 
that  is,  f'(a)=  0.  We  may  determine  whether  f'(x)  passes  from 
positive  to  negative  values  or  vice  versa  by  examining  the  char- 
acter of  its  derivative  in  the  neighborhood  of  x  =  a.  In  case  of  a 
maximum,  f'(x)  is  first  positive,  then  zero  at  x=a,  and  finally 
negative  for  values  following  a.  In  other  words,  as  f(x)  passes 
through  the  maximum  value,  fix)  decreases.  Hence,  by  Art.  34, 
f"(x)  must  be  either  negative  or  zero  for  x  =  a.  On  the  other 
hand,  as/(#)  passes  through  a  minimum, /'(#)  is  first  negative, 
then  zero,  and  finally  positive ;  that  is,  f'(x)  is  an  increasing 
function  at  the  point  x  =  a.  Hence  f"(x)  must  be  either  positive 
or  zero  for  x=a.  For  the  case  in  which /'(a)  becomes  zero  while 
f"(a)  remains  different  from  zero,  we  may  therefore  express  the 
condition  for  a  maximum  or  a  minimum  as  follows: 

Theorem  II.  If  f'(a)  =  0,  then  f  (a)  is  a  maximum  or  a  mini- 
mum according  asfu{a)  is  negative  or  positive. 

This  theorem  affords  an  easy  method  of  determining  the  maxi- 
mum or  the  minimum  points  of  a  function.  We  equate  the  first 
derivative  to  zero  and  solve  for  the  real  values  of  the  variable.  If 
upon  substituting  these  values  for  the  variable  in  the  second 
derivative  we  obtain  a  negative  number,  we  have  a  maximum ;  if 
we  obtain  a  positive  number,  we  have  a  minimum. 


134 


DIFFERENTIATION  AND  INTEGRATION    [Chap.  VIIL 


It  may  occur  that  both  fix)  and  f"(x)  vanish  for  the  same  value 
a.  In  this  case  the  last  method  of  testing  the  given  function  for 
maxima  and  minima  fails,  and  we  must  apply  the  first  method. 
Theorem  I  must  also  be  applied  when  fix)  changes  sign  by  pass- 
ing through  infinity,  as  x  passes  through  x  =  a,  or  by  having  a 
finite  discontinuity,  as  in  Fig.  26,  (a)  and  (b)  respectively. 

Y 


Fig.  26. 


The  special  cases  in  which  higher  derivatives  vanish  simul- 
taneously with  fix)  will  be  considered  in  a  later  section,  where  a 
second  proof  purely  analytic  in  its  nature  will  be  given. 

Ex.  1.  Examine  the  function  (x  +  2)"2  (x  —  l)3  for  maximum  and  minimum 
values. 

/(«)  =  (*  + 2)*  (*-l)«, 

/'(a)  =  2(x  +  2)(s  -  l)3  +  3(3  +  2)2  (x  -  l)'2 

=  5<*  +  2)(*-l)«  (»+|). 

From  O  +  2)(x-l)2(5z  +  4)=:0, 

we  have  the  roots  x  —  —  2,  +  1,  —  f .     These  are  critical  values  to  be  examined. 
We  may  test  the  value  x  =  —  2  by  examining  the  original  function  ;  thus 
/(-2)=0, 
/(_  2  +  h)=h\-  2  +  h-  1)8<0, 
/(_  2  -  h)  =  &*(-  2  -  h  -  l)3  <  0. 

Therefore,  since  the  function  is  greater  for  x  =  —  2  than  for  values  of  x  just 
preceding  or  following  it,  it  follows  that  for  x  =  —  2,  the  function  has  a  maxi- 
mum. The  value  x  =  1  we  shall  likewise  examine  by  substituting  x  =  1, 
jc  =  1  -f-  /i,  and  as  =  1  —  h  in  the  original  function. 

/(1)  =  0,  /(l  +  A)  =  (l  +  h  +  2)%»>0, 


Art.  84] 


MAXIMA   AND   MINIMA 


135 


For  this  value,  the- function  has  neither  a  maximum  nor  a  minimum,  although 
/'(x)=0.     For  the  critical  value  x=—  f,  we  may  examine  the  derivative 

For        x  =  —  f  —  h,  f'(x)  is  negative, 

and  for  x  =—  $  +  h,  f'(x)  is  positive. 

Since  f'(x)  changes  from  negative  to 
positive  as  x  passes  through  the  value 
—  f ,  the  function  has  a  minimum  for 
this  value  of  x. 

The  graph  of  this  function  is  shown 
in  Fig.  27. 

Ex.  2.   Examine  x3  —  4 x2  +  5  x  -  2 
for  maxima  and  minima. 

f(x)=xs-4x2  +  5x-2, 

f'{x)=  3  x2-8x+5  =  (8x-5)(x-l),  Fig.  27. 

/"(x)  =  6x-  8. 

Putting  /'(x)  =  0,  the  critical  values  are  seen  to  be  x  =  § ,  x  = 


/"(x)  =  6  x  §  -  8  =  2,  and  for  x  =  1,  /"(x)  = 
minimum  for  x  =  f  and  a  maximum  for  x  =  1. 


1.     For  x  =  f, 

2  ;  hence,  the  function  has  a 


Ex.  3.   Let 


Then 


/(*)  = 


logx 
logx 


(logx)5 


Hence  for  /'(x)  =  0,  log  x 
4F 


r4,5; 


1  or  x  =  e  gives  the  critical  value.  For  x  <  e, 
logx<  1,  and  f'(x)  is  negative,  while  for 
x>e,  logx>l  and  f'(x)  is  positive; 
hence,  /(e)  is  a  minimum  value  of /(x). 

Ex.4.    Examine(x— 4)3 -f  5  for  maxima 
and  minima. 


Here 


whence 


/(*)  = 

/'(X): 


(x-4)t+6, 
2 


-^X 


3(x  -  4) 


Fig.  28. 


For  x=4,  f'(x)=co;  hence  x  =  4  is  a 
critical  value.  For  x  >  4,  /'(x)  is  posi- 
tive, while  for  x  <  4,  /'(x)  is  negative.  Hence  as  x  increases,  /'(x)  changes 
from  negative  to  positive,  and  from  Theorem  I,  /(x)  has  a  minimum  at  x  =  4. 
See  Fig.  28. 


136  DIFFERENTIATION   AND   INTEGRATION    [Chap.  VIII. 

EXERCISES 

Examine  the  following  functions  for  maximum  and  minimum  values. 
1.    y  =  x3  -  10  x2  +  30.  2.    y  =  x4  -  6  x2  +  10. 

3.    y=x(x2-l).  4.    J/  =  x20  -  1). 

Vx  -  1  2       * 

7.   y  =  x+  2  Vl  +  x2.  8.    y  =  ex  +  e~*. 

x 

9.    y  =  xe_x.  10.    y  =  sin  *  +  cos  x,  0  <g  x  <^  —  • 

11.  y  =  2  tan  x  +  sec2  x.  12.  y  =(«  -  2)$  +  6. 

13.  ?/ =  xVax+fr.  14.  y2  =  xi  +  a?. 

15.  y  =  x3  —  x?/.  16.  y  =  x_1  log  x. 

17.  y  =(x+2)^(x-3)2.  18.  y  =  x  -  e*. 

85.  Applications  of  maxima  and  minima.  The  theory  of  maxima 
and  minima  has  many  important  applications  in  geometry, 
mechanics,  and  physics,  a  few  of  which  are  given  in  the  following 
problems.  In  some  of  these  problems  the  function  whose  maxi- 
mum or  minimum  value  is  required  is  given;  in  others,  however, 
the  function  must  be  found  from  the  conditions  stated  in  the 
problem.  Frequently  the  function  derived  will  contain  more  than 
one  variable ;  but  in  such  cases,  the  conditions  of  the  problem  will 
furnish  certain  relations  between  the  variables  by  means  of  which 
all  but  one  can  be  eliminated.  The  function  of  the  single  variable 
thus  obtained  will  be  treated  according  to  the  methods  given  in 
the  preceding  section.  A  careful  study  of  the  following  illustra- 
tive examples  will  give  a  grasp  of  the  proper  method  of  attack. 
Ex.  1.  A  rectangular  channel,  Fig.  29,  carrying  a  given  volume  of  water, 
is  to  be  so  proportioned  as  to  have  a  minimum 
wetted  perimeter  (i.e.  the  part  of  the  perim- 
eter of  the  channel  in  contact  with  water) . 
Determine  the  proportions  of  the  channel. 

Let  x  =  width  of  bottom,  and  y  =  height 
of  water.     The  volume  of  water  is  propor- 
29-  tional    to    the    cross    section;    since    this 

volume  is  to  be  constant,  we  have 

xy  =  C.  (1) 

The  wetted  perimeter  p  is  given  by  the  equation 

p  =  x  +  2y.  (2) 


} 


Art.  85]        APPLICATIONS   OF   MAXIMA   AND   MINIMA 


137 


This  is  the  function  whose  minimum  value  is  required.     It  contains  two  vari- 
ables x  and  y,  but  one  may  be  eliminated  by  means  of  the  first  relation ;  thus, 


X 


whence 


Differentiating  (3),  we  get 


x  +2-^. 
x 


dx 


2C 


(3) 
(4) 


In  order  that  p  shall  be  a  maximum  or  minimum,  we  must  have 

1-^  =  0, 

x2 

whence  x  =  V2C, 

and  y  =  |  V2C"; 

that  is,  x  =  2  y. 


(5) 


To  determine  whether  this  ratio  between  bottom  and  side  gives  a  maximum 

or  minimum,  we  observe  that  i®  is  negative  for  x2  <  2  C  and  positive  for 

dx 

x2  >  2  C.     This  shows  that  p  is  a  minimum  f or  x  =  2  y. 

Ex.  2.   The  deflection  of  a  rectangular  beam  of  a  fixed  length  under  a  given 
load  varies  inversely  as  the  product  of  the  breadth  and 
the  cube  of  the  depth.     From  a  log  a  inches  in  diameter, 
a  beam  is  to  be  cut  of  such  dimensions  as  to  make  the 
deflection  a  minimum. 

Denoting  by  z  the   deflection,    and  by  b  and   h  the 
breadth  and  depth,  respectively,  the  equation 


bhs 


(6) 


Fig.  30. 


expresses  the  law  stated  above. 

There  are  two  variables  in  this  function,  but  one  of  them  may  be  elimi- 
nated by  means  of  a  second  equation  derived  from  the  geometry  of  the  figure, 


viz. 


b2  +  h2 


Combining  (6)  and  (7),  we  have 

z  - 


¥ 


By  differentiation  of  (8) ,  we  obtain 

dz  _  k(i  h2-S  a2) 
dh      h\a2-h2y 


CO 
(8) 

(9) 


138  DIFFERENTIATION   AND   INTEGRATION    [Chap.  VIII. 

This  derivative  takes  the  value  zero  when  4  h2  —  8  a2  =  0,  that  is,  when 

2 
The  corresponding  value  of  b  is,  from  (7) ,  -  •     The  student  may  show  that 

for  these  values  z  is  a  minimum,  not  a  maximum. 

EXERCISES 

1.  Divide  a  number  a  into  two  parts  such  that  the  sum  of  their  squares 
shall  be  a  minimum. 

2.  The  range  of  a  projectile  is  given  by  the  expression  B  =  v°  Sln — ^ ,    in 

9 
which  v0  denotes  the  initial  velocity  and  <f>  the  angle  which  v0  makes  with 
the  horizontal.     For  what  angle  <p  is  the  maximum  range  obtained  ?     What 
is  the  maximum  range  ? 

3.  The  efficiency  of  a  screw  as  a  mechanical  device  is  given  by  the 

formula  E==Hl-h^^ 

h  +  fi 
where  the  constant  /*  is  the  coefficient  of  friction,  and  h  is  the  tangent  of  the 
pitch  angle  of  the  screw.     Find  the  value  of  h  for  which  the  efficiency  is  a 
maximum. 

4.  It  is  desired  to  make  an  open-top  box  of  greatest  possible  volume 
from  a  square  piece  of  tin  whose  side  is  a,  by  cutting  out  equal  squares  from 
the  corners  and  then  folding  up  the  tin  to  form  the  sides.  What  should  be 
the  length  of  the  side  of  the  square  cut  out  ? 

5.  A  roofer  wishes  to  make  an  open  trapezoidal  gutter  of  maximum 
capacity  whose  bottom  and  sides  are  each  4  inches  wide  and  whose  sides 
have  the  same  slope.     What  should  be  the  width  across  the  top  ? 

6.  Determine  the  most  economical  proportions  for  a  cylindrical  tank 
with  flat  heads ;  that  is,  find  the  ratio  of  the  length  to  the  diameter. 

7.  Find  the  minimum  distance  from  the  point  (4,  0)  to  the  parabola 
y2  =  G  x. 

8.  Find  the  minimum  distance  from  the  point  (—6,  0)  to  the  hyperbola 

xy  =  8. 

9.  The  radius  of  curvature  of  the  hyperbola  xy  —  c2  at  the  point  (sci,  y\) 

3 

is  (yi   +  yi  )     (see  Art.  92).     Show  that  this  radius  is  a  minimum  at  the 
2  c2  v  J 

point  (c,  c),  and  find  the  minimum  value. 

10.    Determine  the  right  circular  cone  of  minimum  volume  that  can  be 
circumscribed  about  a  given  sphere. 


Art.  85]  EXERCISES  139 

11.  Given  a  triangle  one  of  whose  vertices  lies  at  the  center  of  a  circle  of 
radius  a.  If  two  sides  of  the  triangle  are  radii  of  the  circle,  show  that  £  a2  is 
the  maximum  area  of  the  triangle. 

Find  the  dimensions  of  the  following  inscribed  figures  that  will  give  maxi- 
mum area. 

12.  Rectangle  in  a  circle  of  radius  r. 

13.  Rectangle  in  ellipse  of  semiaxes  a  and  b. 

14.  Isosceles  triangle  in  circle  of  radius  a. 

Find  the  dimensions  of  the  following  inscribed  solids  in  order  that  the 
volume  shall  be  a  maximum. 

15.  Cylinder  in  right  circular  cone. 

16.  Cone  in  a  sphere  of  radius  a. 

17.  Find  the  dimensions  of  a  conical  tent  that  for  a  given  volume  V  will 
require  the  least  material. 

18.  The  power  P  developed  by  a  Pelton  water  wheel  is  proportional  to  the 
speed  v  of  the  wheel  buckets  and  also  to  the  relative  velocity  c  —  v  with 
which  the  jet  issuing  from  the  nozzle  with  speed  c  strikes  the  buckets ;  that 
is,  P  =  kv(c  —  »),  where  A;  is  a  constant.  With  a  given  value  of  the  jet  speed 
c,  determine  the  bucket  speed  v  that  will  give  maximum  power. 

19.  The  weight  of  hot  gas  passing  up  a  chimney  in  a  given  time  is  given 

by  the  formula                                 ftV0.96T-7\ 
W= - i, 

where  T  and  7\  denote  respectively  the  absolute  temperatures  of  the  gas  and 

T 
the  outside  air,  and  7i   is  constant.     Find  the  ratio  —  for  which  IT  is  a 

maximum.  1 

20.  The  strength  of  a  rectangular  beam  varies  as  the  breadth  and  the 
square  of  the  depth.  Find  the  dimensions  of  the  beam  of  maximum  strength 
that  can  be  cut  from  a  log  14  inches  in  diameter. 

21.  Find  the  beam  of  greatest  stiffness  that  can  be  cut  from  a  log  12  inches 
in  diameter,  knowing  that  the  stiffness  varies  as  the  breadth  and  the  cube  of 
the  depth. 

22.  A  body  of  weight  IT  is  dragged  along  a  horizontal  plane  by  means  of 
a  force  P  whose  line  of  action  makes  an  angle  6  with  the  plane.  The  magni- 
tude of  the  force  is  given  by  the  equation 

p=         W 

H  sin  6  +  cos  6 ' 

in  which  /x  denotes  the  coefficient  of  friction.     Show  that  the  pull  is  least 
when  6  =  arc  tan  /*. 


140  DIFFERENTIATION   AND   INTEGRATION    [Chap.  VIII. 

MISCELLANEOUS   EXERCISES 

Find  ^  for  the  following. 
dx2 

1.   ?/  =  x2sinx.  2.   y  =  exsm(x  +  -Y 

3.    x%  +  y%  =  J.  4.    y- 


5.    If  2/  =  Aeax  +  -Be-1"5,  show  that  ^-a2y  =  0 


Vx2  -  a2 

—  a2u  =  i 
dx2 


6.  If  s  =  C\  sin  to  +  C2  cos  kt,  show  that  —  +  k2s  =  0. 

efc2 

7.  If  w  and  »  are  functions  of  as,  show  that 

Dxn(uv)=uDxnv  +  nDxn-h)Dxu  +  £&Lzi2  Dm«-HD>m*u  +  •■•  +  vZVw. 

^ ! 

Observe  that  the  coefficients  and  symbolic  exponents  follow  the  law  of 
coefficients  and  exponents  in  the  expansion  of  a  binomial.  This  is  Leibnitz' 
theorem.     (Compare  Gibson's  Calculus,  §  68.) 

8.  Using  the  result  of  Ex.  7,  find 

(a)  Dxby  when  y  =  ex  cos  X  ; 
(6)  DJy  when  y  =  x3  log  x. 

9.  If  ^L  =  ±  [§£-*&  +  *£],  find  */=/(x)  knowing  that  ^  =  0 

dx'2      £7  L  2  2  J  dx 

when  x  =  0  and  when  x  =  I ;  also  that  y  =  0  when  x  =  0. 

10.  Find  curves  for  which  at  any  point  the  second  derivative  ^-2  has  the 

dx2 

constant  value  2  a.     Which   of   these   curves  passes  through   the   origin  ? 
Which  has  the  slope  a  at  the  point  (—2,  0)? 

11.  Find  three  successive  integrals  of  the  functions 

(a)  —  —  +  x  +  5.  (6)  eax  —  e~ax.  (c)  sin  (kt  +  e). 

x3 

d2x 

12.  Find  expressions  for  the  acceleration  —  for  the  motions  described  by 

the  equations 

(a)  x  =  cie-a*  +  c2e-0< ; 

(6)  *=(Ci  +  c2«)e-a<. 

d2x 

13.  Find  an  expression  for  —  when  the  equation  of  motion  is 

dt2 

x  =  e  a<(ci  cos  /3«  +  c2  sin  /3£) . 

14.  From  (a)  of  Ex.  12  deduce  the  relation 

^-(<*  +  /3)^  +  a/3x  =  0. 
dt2  dt 


Art.  85]  MISCELLANEOUS   EXERCISES  141 

15.  The  ideal  efficiency  of  a  certain  type  of  water  turbine  is  given  by  the 

•  cu  v2  oh  *4~  tt^ ii^ 

equation  E  — « — — ,  where  u  denotes  the  variable  peripheral 

gh 

speed  of  the  wheel.     Find  the  value  of  u  for  maximum  efficiency,  also  the 

maximum  efficiency. 

16.  Find  maximum  and  minimum  values  (if  such  exist)  of  the  functions 

(a)   ^-2x  +  5  (6)  sin  0(1  + cos  0). 

17.  Find  — \  for  the  curve  given  by  the  parametric  equations 

dx2 

x  =  a(0  —  sin0),  y  =  a(\  —  cos0). 

18.  Find  — \  for  the  conic  section  ax2  +  2  hxy  +  by'2  +  2gx  +  2fy  +  c  =  0. 

dx2 

d29 

19.  Find  the  acceleration  —  from  the  equation 


s  =  L  +  r(l  —  cos  0)  —  VL2  —  r2  sin2  0, 

dd 
knowing  that  —  is  a  constant  w0.     This  is  the  acceleration  of  the  piston  of 

the  steam  engine  mechanism,  L  being  the  length  of  connecting  rod,  r  that  of 
the  crank,  and  w0  the  angular  speed  of  the  crank. 

20.  Show  that  the  maximum  and  minimum  values  of  an  integral  algebraic 
function  occur  alternately. 

21.  The  specific  heat  of  superheated  steam  is  given  by  the  equation 
c  =  a  +  pT  +  p(l  +  jp\ — — - ,  where  T  denotes  the  absolute  temperature, 

p  the  pressure,  and  a,  /3,  a,  O,  and  n  are  constants.     If  p  is  kept  constant, 
show  that  c  takes  a  minimum  value  for  some  temperature  Tm  and  that  this 

value  is  a  +  n+    pTm. 
n  +  1 

22.  A  rectilinear  motion  is  such  that  the  acceleration  is  given  by  the  ex- 
pression a  =  ef  +  e  l ;  derive  expressions  for  velocity  and  distance.  Show 
that  if  the  particle  starts  from  rest,  the  distance  is  numerically  equal  to  the 
acceleration. 

23.  Find  expressions  for  velocity  and  distance  when  the  acceleration  is 
given  by  a .  =  m  —  nk2coskt.  Determine  the  constants  of  integration  Ci  and 
C2  by  taking  v  =  0  and  s  =  0  when  t  =  0. 


CHAPTER   IX 


CURVES 


86.  Concavity.  In  the  present  chapter  we  shall  discuss  some 
of  the  applications  of  differentiation  to  plane  curves,  and  develop 
principles  that  will  enable  the  student  to  trace  a  given  curve  and 
study  its  properties. 

A  curve  is  said  to  be  concave  upward  at  any  point  if  an  arc  of 
the  curve  containing  the  point  lies  above  the  tangent  to  the  curve 
at  the  point.  It  is  said  to  be  concave  downward  if  the  tangent  lies 
above  the  arc.  Thus  the  curve  shown  in  Fig.  31  is  concave  down- 
ward between  the  points  A  and  B,  and  it  is  concave  upward  from 
Bto  a 


>  X 


Fig.  31. 


The  condition  for  concavity  upward  or  downward  can  be  ex- 
pressed in  terms  of  the  derivatives  of  the  function  represented 
by  the  curve.   When  the  arc  of  the  curve  y  =f(x)  lies  below  the  tan- 

dij 
gent,  as  between  the  points  A  and  B,  tan  cf>,  and  consequently  -~, 

ctx 

decreases  as  x  increases.     On  the  other  hand,  when  the  arc  lies 

142 


Art.  86]  CONCAVITY  143 

above  the  tangent,  -^  increases  with  x.     We  have  therefore  the 
dx 

general  statement: 

A  given  curve  is  concave  upward  or  downward  according  as  the 

derivative  —  is  an  increasing  or  a  decreasing  function, 
dx 

When  the  derivative  -^  is  decreasing,  its  derivative,  namely  the 
dx 

second   derivative,    — -,  must  be  either  negative  or  zero;   if  -f- 
dx2  dx 

is  increasing,  then  the  second  derivative  is  either  positive  or  zero. 

Therefore,  if  — ^  is  different  from  zero,  the  preceding  statement 

is  equivalent  to  the  following : 

A  given  curve  is  concave  upward  if  the  second  deHvative  is  posi- 
tive,  and  concave  downward  if  the  second  derivative  is  negative. 

Ex.    Given  the  curve  y  =  x3  —  x2  +  6,  investigate  its  concavity. 

The  second  derivative,  — ^  =  6  x  —  2,  is  positive  for  values  of  x  >  h  and  is 
dx? 

negative  for  values  of  x  <  |.     Hence,  to  the  left  of  x  =  |  the  curve  is  concave 

downward,  and  to  the  right  of  that  point  it  is  concave  upward. 


EXERCISES 

Test  the  following  curves  for  concavity  upward  or  downward. 
1.   y  =  x3-2x2 +  5.  2.   y  =  secx.  3.    y  =  ^ 


4.   y  =  ay/x  —  a.  5.    y  =  \ (ex  +  e~x) . 

6.  Show  that  a  curve  is  concave  or  convex  toward  the  X-axis  according 
&y 
dx2 

7.  Show  that  the  ellipse  b2x2  +  a2y2  =  a?b2  is  everywhere  concave  toward 
the  X-axis. 

8.  Test  the  curves  y  —  ex  and  y  =  logic  for  concavity. 

9.  Test  the  expansion  curves  given  by  the  general  equation  pmvn  =  C  for 
concavity. 

10.   Show  that  the  curve  ay2  =  xs  has  two  branches,  each  of  which  is  con- 
vex to  the  X-axis. 


144  CURVES  [Chap.  IX. 

87.  Points  of  inflexion.  A  point  at  which  a  curve  having  a 
continuous  slope  ceases  to  be  concave  upward  and  becomes  concave 
downward,  or  vice  versa,  is  called  a  point  of  inflexion  of  the  given 
curve.  The  curve  shown  in  Fig.  31  has  points  of  inflexion  at  A, 
B,  C.  It  is  evident  that  at  such  points  the  curve  crosses  its 
tangent. 

The  definition  suggests  the  analytic  condition  for  a  point  of 
inflexion.  As  has  been  shown,  so  long  as  the  curve  y  =/(#)  is 
concave  upward  the  first  derivative  increases  as  x  increases,  and 
when  it  is  concave  downward  the  derivative  decreases  as  x  in- 
creases. It  follows  then  that,  as  x  passes  through  a  value  for 
which  the  curve  has  a  point  of  inflexion,  the  first  derivative  passes 
from  an  increasing  to  a  decreasing  function,  or  vice  versa.  See 
Fig.  10.  This  is  precisely  the  condition  that  the  derived  function 
shall  have  a  maximum  or  minimum ;  and  to  determine  the  points 
of  inflexion  of  the  curve,  we  need  only  to  examine  the  derived 
function  for  maxima  and  minima.  It  is  therefore  a  necessary  and 
sufficient  condition  for  a  point  of  inflexion  that  the  second  deriva- 
tive shall  change  sign  as  the  independent  variable  passes  through 
the  critical  value.  This  change  occurs  when  the  second  deriva- 
tive passes  through  zero  or  through  infinity;  hence  the  coordi- 
nates of  the  points  of  inflexion  of  the  curve  may  in  general  be 
found  by  solving  the  equations 

f"(x)  =  0,  /"(*)=<», 

and  determining  whether  f"(x)  changes  sign  as  x  passes  through 
the  values  thus  obtained. 

Ex.-  Given  the  curve  whose  equation  isy  =  5xB  +  6x  —  7.     We  have 

/'<*)  =  £«  =  80*, 

which  vanishes  for  x  =  0.  Moreover,  f"(x)  changes  sign  as  x  passes  from 
negative  to  positive  values.  Hence  the  curve  has  a  point  of  inflexion  at 
x  =  0. 

EXERCISES 

Test  the  following  curves  for  points  of  inflexion. 
1.    y  =  2xs-Sx2  +  4x-  6.  2.    y  =  3  x4  +  4x2-z  +  10. 

3.    {y—  3)2  =  a;+5.  4.    y  =  sin  x.  5.    y  =  cotx. 


Art.  87]     ASYxMPTOTES,  RECTANGULAR   COORDINATES  145 

6.    y  =  x2  —  ex.  7.    y  —  ex  —  e~x.  8.    y  =  arc  cot  x. 

9.   y  =  e~*\  10.    *  =  -^™  11.   y  = 


*2  +  3  1  +  6a;2 

12.  Given  a  continuous  curve.  Draw  a  tangent  to  this  curve,  and  through 
some  fixed  point  draw  a  straight  line  which  shall  so  change  as  to  remain  par- 
allel to  the  tangent  as  the  point  of  tangency  changes.  Show  that  as  the 
point  of  tangency  passes  through  a  point  of  inflexion,  the  line  through  the 
fixed  point  changes  the  direction  of  its  rotation.  For  this  reason,  the  tan- 
gent at  a  point  of  inflexion  is  sometimes  called  a  stationary  tangent.  Ex- 
press in  terms  of  the  derivatives  of  the  given  function  the  condition  for  a 
stationary  tangent. 

13.  Find  the  equation  of  a  curve  which  has  a  point  of  inflexion  at  the 
point  (0,  3)  such  that  the  inflexional  tangent  makes  an  angle  of  45°  with 
the  X-axis.     How  many  such  curves  can  be  found  ? 

14.  Show  that  the  curve  y  = has  three  points  of  inflexion  and  that 

1  +x2 

these  points  lie  in  a  straight  line. 

88.  Asymptotes,  rectangular  coordinates.  An  asymptote  to  a 
plane  curve  is  a  straight  line,  lying  partly  within  the  finite  region, 
which  is  the  limiting  position  of  a  tangent  to  the  curve 
as  the  point  of  tangency  recedes  indefinitely  along  an  infinite 
branch. 

Two  conditions  are  necessary  for  an  asymptote :  (1)  The  curve 
must  have  at  least  one  infinite  branch.  Thus  an  ellipse  having 
no  infinite  branches  cannot  have  an  asymptote.  (2)  The  limiting 
position  of  the  tangent  must  lie  partly  within  the  finite  portion 
of  the  plane.  For  example,  the  tangent  to  a  parabola  at  infinity 
is  not  an  asymptote  since  it  lies  wholly  at  infinity. 

There  are  two  general  methods  of  determining  the  asymptotes 
to  a  curve  whose  equation  is  given  in  rectangular  coordinates. 

Method  of  limiting  intercepts.  The  equation  of  the  tangent  at 
the  point  (xlf  y{)  being 

y-y^f'ix^-xi),  (Art.  38) 

the  intercepts  of   the  tangent   on   the  coordinate   axes   are   re- 
spectively: .  ?/, 

Intercept  on  X-axis  =  xx u\'  W 

/  (xi) 

Intercept  on  F-axis  =  yx  —  ^if'i^h)-  (2) 


146  CURVES  [Chap.  IX. 

If  one  or  both  of  the  intercepts  has  a  finite  value  for  a^  =  oo  or 
yl  =  oo ,  the  infinite  branch  has  an  asymptote,  and  the  equation  of 
the  asymptote  may  be  found  from  the  two  intercepts  or  from  one 
intercept  and  the  limiting  value  of  f'(xi).  The  following  ex- 
ample will  illustrate  this  method. 

Ex.  1.   Examine  the  curve  whose  equation  is  yz  =  x3  —  3  x2  for  asymptotes. 
Differentiating,  we  obtain 

Sy2^  =  Sx2-6x, 
dx 

whence  /'  (a*)  -  x^  ~2xi 

From  (1)  and  (2)  we  have 


2/i 


X-intercept  =  xx «3 =  — ^ , 

zi2-2zi     x{*-2x1' 

T-intercept  =  yx  -  ^  -  2  Xl*  ~  -^ 


For  X\  =  oo ,  these  intercepts  are  respectively  1  and  —  1.  Remembering 
that  the  equation  of  a  line  in  terms  of  its  intercepts  a  and  b  is 

5  +  J-l, 

a     b 

we  have  as  the  equation  of  the  asymptote, 

x  —  y  =  1. 

Method  of  substitution.     Let  f(x,  y)  =  0  be  the  equation  of  the 
given   curve,   and   assume  the   equation   of   the   tangent   to   be 

y  =  mx  -f  b.  (3) 

Combining  these  equations,  we  obtain  the  equation 

f(x,  mx  +  b)  =  0.  (4) 

Now  a  tangent  to  a  curve  may  be  regarded  as  the  limiting  posi- 
tion of  a  secant  line  as  the  two  points  in  which  this  line  inter- 
sects the  curve  are  made  to  approach  coincidence.  Hence  if  the 
line  y  =  mx  -f-  b  is  tangent  to  the  curve  f(x,  y)  =  0,  equation  (4) 
must  have  equal  roots,  and  if  the  point  of  tangency  recedes  in- 
definitely these  roots  become  infinite.  The  condition  that  two 
of  the  roots  of  (4)  shall  be  infinite  is  that  the  coefficients  of  the 
two  highest  powers  of  x  in  (4)  shall  vanish.*     If,  therefore,  We 


*  Rietz  and  Crathorne's  College  Algebra,  Art.  111. 


Art.  88]  ASYMPTOTES  147 

equate  these  coefficients  to  zero,  we  can  determine  the  values  of 
m  and  b,  and  consequently  the  equation  of  the  asymptote. 

Ex.  2.    Examine  for  asymptotes  the  curve  whose  equation  is 

x*  —  x3y  +  xy  —  y8  =  0. 
Substituting  y  =  mx  -f  &,  the  resulting  equation  is 

a*(l-m)  +  «*(-»»*-&)  + bs  =  0. 

Equating  the  coefficients  of  x4  and  Xs  to  0,  we  have 
1  -  m  =  0,      m*  +  b  =  0, 
whence  m  =  1,    and     6  =  —  1. 

Substituting  these  values  in  (3) ,  we  have  as  the  equation  of  the  asymptote 

y  =  x—  1. 

If  in  equation  (4)  the  coefficients  of  xn  and  icn_1  contain  both 
m  and  5,  the  terms  of  degree  lower  than  the  (n  —  l)th  do  not  in 
general  affect  the  determination  of  the  asymptotes.  However,  if 
the  coefficient  of  a"-1  is  zero,  or  if  the  value  of  m  obtained  by 
equating  the  coefficient  of  x*  to  zero  is  such  as  to  cause  that  of 
xn~l  to  vanish  also,  we  must  equate  the  coefficient  of  the  next 
lower  degree  to  zero  in  order  to  have  two  equations  from  which 
m  and  b  can  be  uniquely  determined.  This  coefficient,  in  general, 
will  be  of  the  second  degree  in  b,  and  hence  we  have  in  this  case 
two  parallel  asymptotes,  since  for  each  value  of  m  there  are  two 
values  of  b. 

Method  of  inspection.  In  case  the  asymptotes  are  parallel  to 
one  of  the  coordinate  axes,  we  can  often  determine  such  asymp- 
totes by  inspection.  For  this  purpose  the  equation  of  the  curve 
is  written  in  descending  powers  of  x  or  of  y.  It  will  then  have 
either  of  the  following  forms : 

aar  +  (by  +  c)a?-1+  •••  +£  =  0,  (5) 

*f  +  (fix  +  y)tfl~1+  '"  +r  =  0.  (6) 

If  both  a  and  by  +  c  vanish,  then  two  roots  of  (5)  become  infinite, 

and  the  line  by  +  c  =  0  (7) 

is  an  asymptote.     Similarly,  if  in  (6)  both  a  and  f3x  +  y  vanish, 

theline  fix  +  y  =  0  (8) 

is  an  asymptote  to  the  given  curve.     Consequently,  if  the  given 
equation,  when  arranged  in  descending  powers  of  x  (or  y),  does 


148  CURVES  [Chap.  IX. 

not  have  the  term  xn  (or  yn),  then  the  coefficient  of  xn~l  (or  yn~l) 
equated  to  zero  gives  an  asymptote  to  the  given  curve. 

If  in  (5),  a,  b,  and  c  all  vanish,  then  the  coefficient  of  xn~2,  which 
is  in  general  a  quadratic  in  y,  will,  when  equated  to  zero,  deter- 
mine two  asymptotes,  real  or  imaginary.  A  similar  statement 
applies  to  (6). 

Ex.  3.  The  equation  4  x3  +  2  x2y  -  6  xy2  -  x2  +  3  y2  -  1  =  0  is  a  cubic  in 
which  ys  is  absent.     Hence  arranged  in  descending  powers  of  y,  it  takes  the 

form  o  y*  -  (6  x  -  3)y2  +  (2  x2)y  +  (4  x3  -  x2  -  1)  =  0. 

If  we  make  6  x  —  3  =  0,  the  coefficients  of  y*  and  y'2  both  vanish.  Hence  the 
line  6  x  —  3  =  0,  or  x  =  \  is  an  asymptote.  For  x  =  ^,  we  have  also  y  =  —  f , 
so  that  the  asymptote  cuts  the  curve  at  the  point  Q,  —  f). 

Ex.  4.  The  equation  x2y2  -  4  xy2  +  3  x2?/  —  4  x2  -  5  =  0  is  of  the  fourth 
degree,  but  lacks  the  terms  in  x4,  x3,  y4,  and  y*.     It  may  be  arranged  in  the 

f0rmS  0  x4  +  0  x3  +  (y2  -  4)  x2  -  (4  y2  -  3  y)  a;  -  5  =  0, 

0  y4  +  0  ?/3  -f  {x2  -  4  a;)  y2  +  (3  x)  y  -  (4  x2  +  5)  =  0. 
From  the  first  we  have  two  asymptotes, 

y  +  2  =  0,     and    y  -  2  =  0, 

determined  by  placing  the  coefficient  of  a;2  equal  to  zero.  From  the  second, 
the  asymptotes  x  =  0,  x  —  4  =  0  are  obtained. 

The  methods  given  are  sufficient  to  determine  all  the  asymp- 
totes to  an  algebraic  curve.  The  second  method  will  be  found 
most  convenient  in  determining  asymptotes  that  make  an  oblique 
angle  with  the  axes  of  coordinates,  and  the  third  in  finding  asymp- 
totes parallel  to  the  axes.  If  the  equation  of  the  curve  involves 
a  transcendental  function,  the  first  method  may  be  used. 


EXERCISES 

Find  the  asymptotes  of  the  following  curves  : 

1.   y8  =  x2(x-b).  2.   tf  =  tf(*±JL\m 

\x  —  a) 

3.    y2  =      x*      •  4.    £-*  =  l. 

9       2a -x  a2      b2 

5.    y2  (x-2)=  x2.  6.   y3  -  y2  -  x2  +  Xs  =  0. 

7.    x2y2  -  c2  (x2  +  y2)  =  0.  8.    x3  +  xy2  -y2  =  0. 


Arts.  88,  89] 


SINGULAR  POINTS 


149 


9.    x8  —  xy2  +  ay2  =  0. 


8«3 


li.  y3 

13.    x3 


6  x2  +  x\ 

ys  -  x2  +  2  */2  =  0. 


15.   y  +  zy 


0. 


10.  V 

x2  +  4  a2 

12.   a^  =  1. 

14.   x2y  +  xy2  =  8. 
16.   z2?/2  -  z3  -  ?/3  =  0. 


89.  Singular  points.  Certain  points  of  a  plane  curve  may  have 
peculiarities  not  possessed  by  other  points.  Thus  there  may  be  a 
multiple  point,  where  two  or  more  branches  of  the  curve  intersect, 
Fig.  32  (a)  and  (b) ;  a  tacnode,  where  two  branches  of  the  curve 
come  in  contact  and  have  a  common  tangent  at  the  point,  Fig. 


Fig.  32. 


32  (c)  ;  a  cusp,  where  the  two  branches  terminate  at  the  point 
of  contact  and  have  a  common  tangent,  Fig.  32  (d) ;  or  a 
conjugate  point,  the  coordinates  of  which  satisfy  the  equation 
of  the  curve  yet  through  which  no  branches  of  the  curve 
pass.     All  such  points  are  called  singular  points.     At  a  singular 

point  the  derivative  -^  has  two  or  more  values,  real  or  imaginary, 

(XX 

equal  or  distinct.  Geometrically  this  means  that  at  a  singular 
point  the  curve  has  two  or  more  tangents,  though  some  of  these 
may  be  coincident,  and  some  or  all  may  be  imaginary.  The 
character  of  the  curve  at  a  singular  point  depends  therefore  upon 

the  values  of  -*  at  that  point.     A  general  method  of  evaluating 
dx 


150 


CURVES 


[Chap.  IX. 


-*  at  singular  points  cannot  be  given  until  later;  fortunately, 

ClOC 

however,  simple  special  methods  are  sufficient  for. the  cases  that 
ordinarily  arise.     The  following  examples  are  illustrative. 


Ex.  1.     Examine  the  curve  y2  =  x2  —  x4  for  singular  points. 


Forming    the    derivative,   we   have 


dy  _  2  x  —  4  x3  _  2  x  —  4  xa 


dx 


2y 


For 


2xVl  -x1 


0  <  x  <  1,  or  for  —  1  <  x  <  0,  every  value  of  x  gives  two  distinct  values  of  y 

and  two  corresponding  determinate  values  of  -£.    For  x  =  0,  there  is  one 

dx 

value  of  y,  namely  y  =  0,  showing  that  two  branches  of  the  curve  pass 

through  the  origin.     But  for  x  =  0,  -^  takes  the  indeterminate  form  -  and 

dx  0 

this  must  be  evaluated.    Using  the  method  of  Art.  15,  we  have 


2  x  —  4  x3 


r    l-2x2 

L  -  -=±1. 


tf  =  02xVl-X2       a?  =  0Vl-X2 

Hence  at  the  origin  -^  =  ±  1  and  the  tangents  to  the  two  branches  are 
dx 

respectively  y  =  x  and  y  =  —  x.     Evidently  the  curve  lies  between  the  limits 

x  =  1  and  x  =—  1.     The  student  may  draw  the  curve. 

In  relatively  few  cases  can  tjie  indeterminate  expression  for  -^ 

dx 

be  evaluated  by  simple  methods  as   in   Ex.  1.     Generally  the 

method  shown  in  the  following  example  is  effective  and  is  easily 

applied. 

Ex.  2.     Examine  for  singular  points   the  curve  given  by  the  equation 
x4  -  4  x2y  +  if  =  0. 

In  this  case  more  than  one  branch  of  the  curve  passes  through  the  origin  ; 
therefore  the  origin  is  a  point  to  be  examined.     Substitute  y  =  mx  in  the 
given  equation.     The  result  is  an  equation  in  x, 

x3  (x  -  4  wi  +  m3)  =0,  (1) 

which  has  three  roots  each  equal  to  zero.  Hence 
there  are  three  branches  passing  through  the  origin, 
that  is,  the  origin  is  a  triple  point.  From  (1)  we 
have  also  the  equation 

x  -  4  m  +  m*  =  0,  (2) 

which  gives  the  relation  between  the  slope  m  of  a 
secant  line  y  —  mx  that  cuts  the  curve  at  the  origin 
and  in  a  second  point  whose  abscissa  is  x.     Since 
(2)  is  a  cubic  in  m  there  are  three  such  lines.     As  x  approaches  zero  the 


Art.  89] 


SINGULAR   POINTS 


151 


points  of  intersection  approach  coincidence  and  each  secant  line  approaches 
as  a  limiting  position  a  tangent  at  the  origin.  Therefore  putting  x  =  0  in  (2) 
and  solving  the  resulting  equation  m3  —  4  m  =  0,  we  get  in  the  three  roots 
m  =  0,  2,  and  —  2  the  slopes  of  the  three  tangents  to  the  three  branches 
passing  through  the  origin.     The  curve  is  shown  in  Fig.  33. 

Ex.  3.     Examine  for  singular  points  the  curve  ay  =  (x  —  by. 

In  general  y  has  two  values  for  any  value  of  x  <  b,  but  for  x  =  b,  y  has 
the  single  value  0 ;  hence  the  point  (6,  0)  is  to  be  examined.  For  the  sake 
of  convenience  let  the  origin  be  shifted  to  the  point 
(&,  0).  Substituting  x—b  —  x  and  squaring  to  re- 
move the  radical,  we  have  a2y2  —  x,b  —  0.  Letting 
=  mx' 


i*y2 

;',  we  obtain  the  equation  in  x', 
x'2  {a2m2  -  x  3)  =  0, 
which  has  two  zero  roots,  showing  that  two  branches 
pass    through     the     origin.        From     the     equation 


a'm' 


0,  we  have  f or  x  =  0,  a2m2  =  0,  an  equa- 


>X 


tion  in  m  with  equal  zero  roots.  It  follows  that  the 
branches  have  a  common  tangent  y  —  0.  For  nega- 
tive values  of  x\  that  is,  for  x<b,  y  takes  imaginary 

values ;  hence  the  two  branches  terminate  at  the  point  of  contact,  and  this 
point  (6,  0)  is  therefore  a  cusp,  Fig.  34. 

Ex.  4.     Examine  the  curve  x*  —  3  x2  —  3  y%  +  yz  =  0. 

The  origin  is  evidently  a  point  to  be  ex- 
amined.     Putting  y  =  mx,  the  result  is 

&  _  3  aj2  __  s  W2X2  +  mZjfi  =  o, 
or  x2  (x  -  3  -  3  m2  +  mzx)  =  0. 

It  follows  that  two  branches  (real  or  im- 
aginary) pass  through  the  origin.  To  de- 
termine the  tangents  at  the  origin,  we  place 
x  —  0  in  the  equation  x  —  3  —  3  m2  +  mzx  =  0. 
The  result  is  3  m2  +  3  =  0,  from  which 
m=±  V—  1.  Since  the  tangents  are  imagi- 
nary no  real  branches  pass  through  the 
origin  and  therefore  the  origin  is  a  conjugate  point.  The  curve  is  shown  in 
Fig.  35. 


KX 


EXERCISES 

Examine  the  following  curves  for  singular  points 
1.  as*  -  2  axy2  +  2  ay*  =  0.  2.   y2  = 


3.    y  = 


2a  —  x 
4.   aV2  =  a2x*  -  x6. 


Vx2 


152  CURVES  [Chap.    IX. 

5.   x(x2  +  y2)  +  a(x2  -  y2)  =  0.  6.   xs  -  3  axy  +  ys  =  0. 

7.  Show  that  the  leinniscate  (x2  +  y2)2  =  a2(x2  —  y2)  has  a  double  point 
at  the  origin,  and  find  the  tangents  at  this  point. 

8.  Show  that  the  curve  y2  =  xs  —  4  x2  has  a  conjugate  point  at  the  origin. 

2  2  2 

9.  Show  that  the  curve  Xs  +  y*  —  a3  has  a  cusp  at  each  of  the  points 
(a,  0),  (-  a,  0),  (0,  a),  and  (0,  -  a). 

10.  Find  the  equation  of  a  curve  that  has  a  double  point  at  the  origin 
with  tangents  y  =  x,  and  y  =  2  x. 

11.  Examine  the  curve  y  —b  =(x  —  a)  y/x  for  singular  points. 

12.  The  curve  y2  =  a£2  +  &«*  has  a  double  point  at  the  origin.  Find  the 
tangents  at  the  origin  by  evaluating  -^  for  x  =  0. 

90.  Curve  tracing.  It  is  frequently  desirable  to  determine  the 
general  form  of  the  graph  of  a  given  function.  The  direct  method 
of  tracing  a  curve  is  to  determine  simultaneous  values  of  x  and  y 
from  the  given  equation  y  =f(x)  and  plot  the  points  thus  found. 
However,  by  a  careful  study  of  the  derivatives  of  the  function, 
the  labor  of  this  direct  method  can  be  largely  avoided,  and  the 
general  course  of  the  graph,  which  is  all  that  is  required,  can  be 
found.  In  tracing  a  curve  the  student  should  proceed  somewhat 
as  follows : 

1.  Examine  the  equation  for  symmetry.  If  only  even  powers 
of  x  appear,  the  curve  is  symmetrical  with  respect  to  the  y-axis ; 
if  only  even  powers  of  y  appear,  it  is  symmetrical  with  respect  to 
the  X-axis ;  while  if  the  equation  remains  unchanged  when  x  and 
y  are  replaced  by  —  x  and  —  y  respectively,  the  curve  is  sym- 
metrical with  respect  to  the  origin. 

2.  Find  the  points  in  which  the  curve  crosses  the  X-  and  Y"-axes. 

3.  Find  the  finite  values  of  x  (or  of  y)  for  which  y  (or  x)  be- 
comes infinite. 

4.  Find  the  slope  of  the  curve  by  means  of  the  first  derivative 
and  note  the  turning  points. 

.5.  By  an  examination  of  the  first  or  second  derivative  deter- 
mine whether  the  curve  is  concave  upward  or  downward,  and 
find  the  points  of  inflexion. 

6.   Examine  the  curve  for  asymptotes. 


Art.  90]  CURVE   TRACING  153 

7.  In  some  cases  it  may  be  advisable  to  examine  the  curve  for 
singular  points. 

Having  determined  these  characteristics  of  the  curve,  it  is  easy 
in  most  cases  to  sketch  the  graph.  If  a  minute  study  of  the  form 
of  the  curve  at  any  particular  point  is  desired,  it  is  well  to  trans- 
form the  equation  so  that  the  point  in  question  becomes  the  ori- 
gin ;  then  we  need  only  consider  the  nature  of  the  given  curve 
in  the  neighborhood  of  the  origin.  In  determining  the  nature  of 
the  curve  in  the  neighborhood  of  the  origin,  the  terms  of  the 
lowest  degree  in  x  and  y  are  in  general  most  important,  since  for 
small  values  of  these  variables  the  terms  of  higher  order  are  small 
in  comparison.  Likewise  the  nature  of  the  curve  for  very  large 
values  of  the  variables  is  determined  by  considering  only  the 
terms  of  highest  order  in  the  variables. 

In  the  following  examples  we  shall  consider  a  few  of  the 
simpler  curves,  which  are  frequently  referred  to  in  subsequent 
chapters. 

Ex.  1.  The  curves  y  =  axn.  It  is  assumed  that  n  is  positive  and  either 
an  integer  or  a  fraction.  Eor  n  =  2  and  for  n  =  J,  the  curves  are  ordinary 
parabolas.  If  n  =  3,  we  have  the  cubical  parabola  y  =  ax?.  An  examination 
of  this  equation  discloses  the  following :  (a)  The  curve  passes  through  the 
origin.  (&)  For  positive  values  of  se,  y  is  positive,  and  for  negative  values  of 
x,  y  is  negative  ;  hence  the  curve  lies  wholly  in  the  first  and  third  quadrants, 

and  is  symmetrical  with  respect  to  the  origin,  (c)  -"  =  3  ax2  ;  hence  «the 
slope  is  always  positive  and  increases  as  the  numerical  value  of  x  increases. 

At  the  origin  the  slope  is  zero,     (d)    — ^  =  6  ax,  hence  to  the  right  of  the 

dx2 

F-axis  the  curve  is  concave  upward,  and  to  the  left  of  the  F-axis  it  is  con- 
cave downward,     (e)  At  £=0,  (-^/-  changes  sign  ;  hence  the  origin  is  a  point 

of  inflexion.  From  these  conclusions  it  is  evident  that  the  curve  has  the 
general  form  shown  in  Fig.  36. 

If  n  —  f,  the  equation  becomes  y2  =  a2x3,  and  the  curve  is  the  semicubical 
parabola.     From  the  equation  we  get 

dx     2  dx2     4  Va- 

It  is  readily  seen  that  the  curve  is  symmetrical  with  respect  to  the  X-axis 
and  lies  wholly  to  the  right  of  the  F-axis.  At  the  origin  the  slope  is  zero, 
and  the  two  branches  of  the  curve  have  the  X-axis  as  a  common  tangent ; 


154 


CURVES 


[Chap.  IX. 


the  origin  is  therefore  a  cusp.     There  is  no  point  of  inflexion.     Hence  the 
curve  has  the  general  form  shown  in  Fig.  37. 


^X 


yx 


*~X 


Fig.  36. 
Ex.  2.     The  cissoid  y2  = 


Fig.  37. 


Fig.  38. 


Xs 


The  curve  is  symmetrical  with  respect 

to  the  X-axis.    For  x  =  2  a,  y  becomes  infinite  ;  and  for  x  >  2  a  or  x  <  0,  y 
is  imaginary  ;  hence  the  curve  lies  wholly  between  the  Y-axis  and  the  line 

x  =  2a,  which  is  an  asymptote.     For  x  =  0,  ^  =  0,  showing  that  the  X-axis 

dx 

is  a  tangent  to  both  branches  at  the  origin.     An  investigation  of  the  second 

derivative  shows  that  there  is  no  point  of  inflexion  and  that  both  branches 

are  convex  towards  the  X-axis.     The  curve  has  therefore  the  general  form 

shown  in  Fig.  38. 

$a* 


Ex.  3.      The  witch  y  = 


The  curve  is  symmetrical  with  respect 


z:  +  4a2 
to  the  Y-axis,     y  has  a  maximum  value  2  a  for  x  =  0,  and  there  are  points 

Y 


Fig.  40. 


of  inflexion  at  x  =  ±2aVS.     The  X-axis  is  an  asymptote.     The  curve  is 
shown  in  Fiff.  39. 


Art.  90] 


CURVE  TRACING 


155 


Ex.  4.    The  lemniscate  p2  =  a2  cos  2  0, 
or  (x2  +  y2)2  =  a2(x2-y2). 

The  curve  is  symmetrical  with  respect  to  both  axes,  and  crosses  the  X-axis 
at  x  =  ±  a.  For  values  of  0  from  \  ir  to  f  ir  and  from  f  w  to  \  w,  p  is  imagi- 
nary. The  origin  is  a  double  point,  the  tangents  having  the  slopes  —  1  and 
1,  respectively.     The  curve  has  the  form  shown  in  Fig.  40. 

X  X 

Ex.  5.  TJie  catenary  y  =  -  (ea  +  ca)  is  the  curve  assumed  by  a  flexible 
A 
cord  of  uniform  weight  suspended  trom  two  fixed  points.  The  curve  is  sym- 
metrical about  the  T-axis,  and  cuts  the  y-axis  at  a  distance  a  above  the 
origin.  The  second  derivative  is  positive  for  all  values  of  x,  hence  the  curve 
is  concave  upwards  and  has  no  point  of  inflexion.     See  Fig.  41. 


Fig.  42. 


Ex.  6.  The  cycloid  is  the  curve  described  by  apoint  on  the  circumference 
of  a  circle  which  rolls  on  a  straight  line.  Let  a  denote  the  radius  of  the 
circle,  and  0  the  angle  PCM,  Fig.  42,  subtended  by  the  arc  PM {=■  OM). 
Then  we  have  for  the  coordinates  of  P,  x  =  a(d  —  sin  0),  y  =  a(l  —  cos  6). 
From  these  equations,  we  readily  obtain 

dy  _     sin  d         d?y 1_ 

dx  ~  1  —  cos  0'    dx2  ~      a(l  -  cos  0)2 

The  curve  has  a  turning  point  at  x  —  ira,  and  since  — ^  is  always  negative, 

dx2 
the  curve  is  everywhere  concave  downward. 

Ex.  7.  The  astroid,  or  hypocycloid  of  four  cusps,  x'5  +  y*  =  a?  is  a  curve 
described  by  a  point  on  the  circumference  of  a  circle  of  radius  \  a  rolling 
within  the  circumference  of  a  circle  of  radius  a.  See  Fig.  43.  The  student 
may  find  the  derivatives  and  by  their  aid  study  the  curve  for  slope,  concav- 
ity, etc. 

Ex.  8.  The  cardioid  />  =  2a(l  —  cos0).  The  curve  is  closed,  and  p  is 
finite  for  all  values  of  0.  Since  cos  0  =  cos  (—0),  the  curve  is  symmetrical 
with  respect  to  the  initial  line  OX.     From  the  given  equation,  we  get 


tan  \p  =z  p 


dd     1  —  cos  0 


whence 


sin0 


=  tan  \  0, 


(Art.  40) 


156 


CURVES 


[Chap.  IX. 


For  6  =  0,  \p  =  0,  that  is,  OX  is  a  tangent  at  the  origin  ;  for  d 


r)f  =  iir, 


and  for  d  =  tt,  \p  =  \  ?r.     For  the  limits  of  p  we  have  p  =  0  when  0  =  0,  />=  4  a 
when  0  =  v.     See  Fig.  44. 


Fig.  43. 


Fig.  44. 


EXERCISES 

The  student  should  trace  the  following  curves  carefully  and  preserve  the 
graphs  for  future  reference. 

1.  The  group  of   curves  xmy  =  G{m  positive).     These  are   the  curves 
which  represent  the  laws  of  expanding  gases. 

2.  The  curve   x-  +  y*  =  a?.     Show   that    this  curve   is    the  ordinary 
parabola. 

3.  The  folium  of  Descartes,  Xs  +  y8  —  3  axy  =  0. 

4.  The  exponential  curve  y  =  ex.       5.    The  logarithmic  curve  y  =  log  x. 
6.    The  logarithmic  spiral  p  =  eae.      7.    The  spiral  of  Archimedes,  p=a0. 


8.   The  curve  p  =  a  sec2 


9.    The  parabolic  spiral  p2  =  a20. 


10.    Rankine's  equation  for  columns,  y  = 


1  +  bx* 

11.  The  probability  curve  y  =  ke-"*2,  a  <  0.     Find  turning 'points  and 
points  of  inflexion. 

12.  The  lituus  pH  -  a2. 

13.  The  curves  p  =  a  sin  nd,  p  =  a  cos  nd.     (Give  n  values  1,  2,  3,  4.) 

14.  Show  that  the  cardioid  is  described  by  a  point  on  the  circumference 
of  a  circle  of  radius  a  which  rolls  upon  a  fixed  circle  of  the  same  radius. 


Art.  91] 


CURVATURE 


157 


91.  Curvature.  If  a  point  moves  along  a  plane  curve  as  EF, 
Fig.  45,  the  direction  of  motion  at  any  point  is  the  direction  of 
the  tangent  to  the  curve. 
The  direction  of  the  tan- 
gent continually  changes, 
and  a  comparison  of  this 
change  of  direction  with 
the  distance  traversed  by 
the  point  leads  us  to  the 
idea  of  curvature.  Thus, 
as  the  point  moves  from 
P  to  Q,  the  tangent  turns 
through  the  angle  A<£,  and 
we  say  the  curvature  of 
the  arc  PQ  is  large  or  small  according  as  the  angle  Acf>  is  large  or 
small. 

Denoting  by  As  the  length  of  the  arc  PQ,  the  ratio  — *  is  de- 

fined  as  the  mean  curvature  of  the  arc  PQ.  Provided  the  curva- 
ture is  constant,  this  quotient  is  the  curvature  at  all  points  be- 
tween P  and  Q ;  but  if  the  curvature  is  not  constant,  then  the 
curvature  at  P  is  defined  as  the  limit 


As  =  0  As  ds 


(1) 


The  curvature  of  the  curve  y  =f(x)  at  the  point  (a^,  ft)  may  be 
expressed  in  terms  of  the  derivatives  f'(x)  and  f"(x).  We  have 
the  fundamental  relation 

tan  0  =f\x),     <j>  =  arc  tan  f(x), 
whence  by  differentiation 


Furthermore,  we  have 


i  +  [/'(*)]2    i  +  [/'(*)T 


Dj*  =  y/l+[f{x)y* 


(See  Art.  47) 


*  A  derivation  of  this  relation  is  given  in  Art.  108.     For  the  present,  the 
student  may  make  use  of  Eq.  (1),  Art.  47. 


158  CURVES  [Chap.  IX. 

From  these  equations,  we  obtain  by  division 


At  the  point  (x1}  y^,  therefore,  the  curvature  is 

f"(vi) 


(2) 


(3) 


92.  Radius  of  curvature.  Center  of  curvature.  Suppose  that 
PQ,  Fig.  45,  is  an  arc  of  a  circle.  Let  normals  be  drawn  at  P 
and  Q  intersecting  at  M\  then  MP=  MQ  is  the  radius  of  the  arc, 
and  angle  PMQ  =  A<£.     Denoting  the  radius  by  r,  we  have 


rA<£  = 

=  As; 

A<£_ 

As" 

1. 

-    5 
r 

whence  — -  =  -;  (1) 

As      r 

that  is,  the  curvature  of  a  circle  is  constant  and  is  the  reciprocal 
of  the  radius. 

At  any  point  of  a  curve  y  =f(x),  conceive  a  circle  drawn  tan- 
gent to  the  curve,  and  suppose  it  to  have  the  same  curvature  as 
the  curve  at  that  point.  This  circle  is  called  the  circle  of  curva- 
ture ;  its  radius,  the  radius  of  curvature ;  and  its  center,  the  center 
of  curvature  of  the  curve  at  the  given  point.  Since  the  circle  and 
curve  have  the  same  curvature  at  the  point  in  question,  the  radius 
of  curvature  R  must  be  the  reciprocal  of  this  curvature ;  that  is, 

11  =  ^-. 
d<f> 

In  terms  of  the  derivatives  involved,  the  formula  for  the  radius 
of  curvature  at  the  point  (xx,  y{)  becomes,  therefore, 

R /»(*!)  *  (2) 

By  hypothesis,  the  curve  and  the  circle  of  curvature  at  the 
given  point  have  a  common  tangent;  hence  the  radius  of  curva- 
ture has  the  direction  of  the  normal  to  the  curve. 


Art.  92]  RADIUS  AND   CENTER   OF   CURVATURE 


159 


It  is  customary  in  the  case  of  single-valued  functions  to  regard 
R  positive  or  negative  according  as  the  curve  is  concave  upward 
or  concave  downward.  To  establish  this  convention,  it  is  neces- 
sary to  take  the  positive  value  of  the  radical  in  the  numerator. 
The  sign  of  R  then  depends  upon  that  of  f"(x),  and  the  required 
result  follows  from  Art.  87.  In  many  cases  it  is  the  numerical 
value  of  R  alone  that  is  of  importance. 

d2v 
As  we  have  seen,/"(#)  =  — ^  changes  sign  at  a  point  of  inflexion ; 

hence  R  also  changes  sign  at  such  a  point.  This  property  might 
have  been  used  as  the  definition  of  a  point  of  inflexion.  Thus  a 
point  of  inflexion  is  a  point  at  which 
the  radius  of  curvature,  and  conse- 
quently the  curvature  itself,  changes 
sign. 

The  coordinates  of  the  center  of 
curvature  may  be  found  as  follows  : 
If  (m,  n),  Fig.  46,  is  the  center  of 
curvature  corresponding  to  the  point 
(xd  Vi)  °f  the  curve,  we  have  from 
the  figure, 

x1  —  m  =  R  sin  <j>,   n  — 


Fig.  46. 


yx=  R cos  <j>. 


But 

whence     cos  <f> 


1 


Vl+L/'Caa)]* 


tan  <£  =f(x1), 
,    sin  <f>  = 


/'(*,) 


Vl +[/'(*,)  j 


Using  these  expressions  for  sin  <£  and  cos  <f>,  and  the  expression 
for  R  given  by  (2),  we  obtain  after  reduction 


m  =  oci  — 
n  =  yx  + 


/'(^iMi+r/'(^i)]2S 

/"(<Bl) 

l+[/'(a?i)]2 


(3) 


Ex.   Find  the  radius  of  curvature  of  the  equilateral  hyperbola  xy  =  c2  at 
the  point  (xi,  yi);  also  at  the  point  (c,  c). 


We  have 


X  X\2  Xi3 


160  CURVES  [Chap.  IX. 

Substituting  these  values  in  (1),  we  get 

L  X,\\      _(X12+Vl2)* 


B 


2  c2  2  c2 


At  the  point  (c,  c),  therefore,  B  =  (2  c  ^  =  cV2. 

2  c2 

o 
Also,  at  this  point  /'(as)  =  —  1,  and  /"(as)  =-■     Substituting  in  (3),  we  find 

c 

for  the  coordinates  of  the  center  of  curvature, 

m  =  n  =  2  c. 


EXERCISES 

Derive  general  expressions  for  the  radius  of  curvature  of  each  of  the  fol- 
lowing curves : 


1. 

y2  =  2  aas. 

2. 

ayB  =  as2  +  &• 

3. 

x2     y2_l 
a2      b2 

4. 

y  =log(aj  +  a). 

5. 

y  =  log  sec  x. 

6. 

2              2             2 

Xs  +  y3  =  a5. 

7. 

y  =  -(e«  +  e~«). 

8. 

x2      y2  _  . 
a2      62_ 

9. 

3 

y3  =  x^  —  5. 

10. 

yz  =     

2a—  x 

11. 

Vas  +  Vy  =  2  Vc". 

For  the  following  curves,  find  the  radius  of  curvature  and  the  coordinates 
of  the  center  of  curvature  at  the  points  indicated. 

12.    y2  =  lQx,  at  as  =  5.     13.  xy  =  30,  at  (3,  10).     14.  y  =  cos as,  at  (0,  1). 

15.  Find  the  radius  of  curvature  of  the  probability  curve  y  =  ke  ax2  for 

3  =  0. 

16.  Derive  an  expression  for  the  curvature  of  the  parabola  y  =  ax2 +6as+c, 
and  show  that  the  curvature  is  maximum  at  the  vertex. 

17.  Find  the  point  of  maximum  curvature  of  the  exponential  curve  y  =  ex. 

93.    Radius  of  curvature,  parametric  representation.     If  a  curve 
is  given  by  parametric  equations,  as 

we   may   obtain   an   expression  for   the   radius  of  curvature  as 
follows : 


(1) 


Art.  93]  RADIUS   OF   CURVATURE  161 

We  have  the  general  formula 

dy 
dy  =  dO 
dx     dx 

dO 

from  which,  by  the  aid  of  Art.  31,  we  obtain 

dx  d2y      dy  d2x 
d2y^dddti2~d6dFdO 
dx2  ~         fcI^\2         dx 

W 

dx  dry  _  dy  d2x 

do'cr2~dede2 

dx^6 
d$ 


(2) 


Substituting  these  expressions  in  formula  (2),  Art  92,  we  obtain 

KS1 + 


(dy 
\d0 


dx  d2y     dy  d2x 
d0d62~d$de2 

Ex.    Find  the  radius  of  curvature  of  the  cycloid 

x  =  a(d  —  sin  0), 

y  =  «(1  —  COS0). 

By  successive  differentiations,  we  obtain 

^  =  a(l-cos0),         -^  =  asin0, 
dd        K  dd 

—  =  a  sin  9,  ^  =  a  cos  6. 

dff?<  dff2 

Substituting  these  values  in  (3),  we  find 

a2  (cos  6>-l)  2 

If  a  curve  is  given  in  polar  coordinates,  we  have 
x  =  p  cos  0,   ?/  =  p  sin  0. 


(3) 


162  CURVES  [Chap.  IX. 


By  means  of  these  relations  (3)  is  reduced  to  the  form 

(5) 


-(% 


*+21*b\*-~*£ 


R  = 

Using   the  functional    symbols  for   the    derivatives,  the    radius 
of  curvature  of  the  curve  p  =  F{6)  at  the  point  (plf  0^)  is 

R  = ^1^L     v  lM (6) 

Pl2+2[F>(61)Y-p1F>X0l) 


EXERCISES 

1.  Find  the  radius  of  curvature  of  the  ellipse  from  the  equations 

X  =  a  cos  0,        y  =b  sin  0. 

2.  Find  the  radius  of  curvature  of  the  hypocycloid  from  the  equations 

x  =  a  cos3  0,         y  =a  sin3  0. 
Derive  general  expressions  for  the  radii  of  curvature  of  the  curves  given 
by  the  following  equations  in  polar  coordinates. 

3.  p  =  ad.  4.   p  =  a(sin  0  +  cos  6). 

a 

5.    />  =  asin3--  6.    p  =  ea0. 

7.  Find  the  maximum  radius  of  curvature  of  the  cardioid  p  =  2  a(l  —  cos  6). 

8.  Find  the  minimum  radius  of  curvature  of  the  lemniscate  p2  =  a2  cos  2  0. 

9.  From  the  answer  to  Ex.  1  show  that  the  radius  of  curvature  of  the 
ellipse  is  greatest  at  the  extremity  of  the  minor  axis  and  least  at  the  extremity 
of  the  major  axis. 

94.  Roulettes  and  involutes.  Suppose  a  plane  curve,  as  AB, 
Fig.  47,  to  roll  without  slipping  on  a  fixed  curve  MN\  for  this 
purpose  we  may  regard  the  curves  as  the  boundaries  of  two  disks. 
Points  E,  F  on  the  rolling  curve  describe  curves  e,  f  on  the  fixed 
plane.  Curves  generated  in  this  manner  are  called  roulettes. 
Cycloids  and  trochoids  are  examples  of  roulettes  in  which  the 
rolling  curve  is  a  circle.     (See  Art.  90.) 

If  the  rolling  curve  is  replaced  by  a  straight  line,  Fig.  48,  the 
roulettes  are  called  involutes  of  the  fixed  curve;  thus,  curves  e 
and /are  involutes  of  the  curve  MN.  Evidently  a  curve  has  an 
infinite  number  of  involutes. 


Arts.  94,  95] 


ROULETTES  AND   INVOLUTES 


163 


AVe  may  also  consider  the  involute  to  be  generated  by  a  point 
on  a  flexible  thread  or  cord  wrapped  around  the  fixed  curve.  As 
the  cord  is  unwrapped,  any  point  of  it  generates  an  involute. 


Fig.  47. 


Fig.  48. 


Certain  properties  of  involutes  are  evident  from  the  manner  in 
which  they  are  generated. 

1.  Any  normal  to  an  involute  is  a  tangent  of  the  fixed  curve. 
This  property  may  be  seen  from  purely  mechanical  principles. 
Keferring  to  Fig.  48,  the  point  of  contact  P,  considered  as  a  point 
of  the  rolling  line  AB,  has  at  the  instant  of  contact  no  motion  in 
the  direction  of  the  line,  since  by  hypothesis  there  is  no  slipping. 
Furthermore,  since  the  curves  remain  in  contact,  P  can  have  no 
motion  perpendicular  to  AB.  It  follows  that  the  point  P  of  AB 
is  at  rest,  and  the  curve  as  a  whole  is  rotating  about  P  as  a  center. 
Points  E  and  P  are  therefore  moving  in  directions  perpendicular 
to  the  lines  PE  and  PF  respectively.  Now  the  point  E  moves  in 
the  direction  of  the  tangent  to  the  curve  it  describes ;  hence  the 
line  PE  is  the  normal  to  the  curve  e  at  E,  and  likewise  PF  is 
normal  to  curve/  at  F.  In  the  case  of  the  involute,  Fig.  48,  the 
normals  PE  and  PF  coincide  with  the  rolling  line  AB,  which  is 
tangent  to  MN\  hence  the  normal  of  the  involute  is  tangent  to 
the  curve. 

1.  Two  involutes  of  the  same  curve  intercept  a  constant  distance 
on  their  common  normal,  This  follows  at  once  from  the  manner 
in  which  the  involute  is  generated.  Because  of  this  property, 
involutes  are  sometimes  called  parallel  curves. 

95.  Curvature  of  involutes.  Let  PQ,  Fig.  49,  be  one  position 
of  the  rolling  straight  line,  P  being  the  point  of  contact,  and  Q  sl 


164 


CURVES 


[Chap.  IX. 


point  on  the  involute.     Let  m,  n  denote  the  coordinates  of  the 
variable  point  Pon  the  fixed  curve  MNf  and  x,  y  those  of  the  point 


lY 


(*,y) 


Q  on  the  involute.  From 
the  geometry  of  the  fig- 
ure, we  have  at  once 

(m— #)tan  <£  =  n  — y.  (1) 

Since  QP  is  tangent  to 
MN,  we  have 


tan  cf)  — 


dn  m 
dm 


Fig.  4U. 


dx 


but    since    QP    is    the 

*  X    normal  to  the  involute 

at  the  point  Q,  we  have 


also  tan  <l>  =  —  — .     Substituting  this  value  in  (1)  we  obtain 
dy 

(m—x)  +  (n  —  y)-^  =  0. 

(XX 


(2) 


Differentiating  (2)  with  respect  to  n,  we  have 


dm      dx 
dn      dn 


But  since 

this  equation  reduces  to 


dy\dy      ,  . 


dn 


dm 
dn 


dx      dy  dy      . 
dn      dn  dx      ^ 

dn 


dy 
dx' 


i,)     dn 


and  multiplying  through  by  — ,  we  obtain  finally, 

(XX 


whence 


-'-( 


dx        v       9Jda? 


1  + 


n  =  y  + 


d2y 
dx2 


(3) 


Art.  96] 


EVOLUTE    OF   A   CURVE 


165 


Substituting  this  value  of  n  in  (2),  we  have 


dxl       \dxj  J 


m  =  x 


d2y 
cte2 


(4) 


Comparing  (3)  and  (4)  with  (3)  of  Art.  92,  it  appears  that  the  point 
P  is  the  center  of  curvature  and  PQ  is  the  radius  of  curvature 
at  the  point  Q  of  the  involute. 

It  is  not  true  of  roulettes  in  general  that  the  point  of  contact 
of  the  rolling  curves  is  the  center  of  curvature  of  the  roulette. 
Thus  in  Fig.  47,  the  center  of  curvature  of  e  at  the  point  E  lies 
somewhere  on  EP  or  EP  produced,  but  not  at  P.  Only  in  the 
case  of  the  involute  does  the  center  of  curvature  coincide  with 
the  center  of  rotation. 


96.  Evolute  of  a  curve.  From  the  preceding  article  it  appears 
that  a  given  curve  is  the  locus  of  the  centers  of  curvature  of  each 
of  its  involutes.  When  two  curves  Cx  and  C2  are  so  related  that 
Ci  is  an  involute  of  C2,  we  call  C2  the  evolute  of  Cv  Curve  C2  may 
have  other  involutes  than  Ci,  but  Ci  has  only  the  one  evolute  C2. 

Two  properties  of  the  evolute  follow  from  the  manner  of  de- 
scribing the  involute  as  a  roulette,  namely : 

1.  Any  normal  to  the  curve  Ci  is 
tangent  to  the  evolute  C2. 

2.  The  difference  between  two  radii 
of  curvature  of  a  given  curve  is  equal 
to  the  arc  of  the  evolute  between  the 
points  of  contact  of  the  radii  ivith  the 
evolute. 

Thus  in  Fig.  50,  we  have 

P2E2-P1E1  =  arc  PXP2 


Fig.  50. 


To  obtain  the  equation  of  the  evolute  of  a  given  curve,  we  com- 
bine the  equation  of  the  curve 

y=m  (l) 


166 


CURVES 


[Chap.  IX. 


with  the  equations 


n  =  2/  + 


_\dx[ 
dx2 


dx\ 
m  —  x — 


^fi  +  (*£ 


\dx 


dx2 


(2) 


(3) 


which  give  the  coordinates  m,  n  of  the  center  of  curvature.  If 
x,  y,  and  the  derivatives  be  eliminated  between  these  equations, 
the  result  will  be  a  relation  between  m  and  n,  the  variable  coor- 
dinates of  the  e volute.  Various  special  expedients  may  be  used 
in  the  elimination. 

Ex.  1.    Find  the  evolute  of  the  rectangular  hyperbola  xy  =  — . 


We  have 


Hence 


Therefore 


and 


dy 
dx 


2x2 


d2y  =  a? 
dx2     xs ' 


2x 


4  a:4  +  a4 
4a2x 


x  + 


4x* 


4 

^  +  3a4 

4  a2x 

12  x4  +  ai 

8x* 


8x3 


ra  +  n 


8x*  +  12  x4a2  +  6  x2a*  +  «6 

8a¥ 


\    2  ax     J. 


2  ax 
2x2\ 


(a2-2x2\ 

m  —  n=a[ ] 

\     2  ax     J 

(m  +  riy  —  (m  —  ny 
which  is  the  desired  equation  of  the  evolute. 


Finally 


2d1 


EXERCISES 
Find  the  equations  of  the  evolutes  of  the  following  curves. 

1.  The  circle  x2  +  y2  =  a2. 

2.  The  parabola  y2  =  4px. 


3.  The  ellipse  ^  +  £- 
*      a2     b2 


1. 


Art.  96]  MISCELLANEOUS   EXERCISES  167 

4.  Find  the  e volute  of  the  ellipse  using  the  parametric  equations 
x  —  a  cos  d,  y  =  b  sin  6. 

5.  Find  the  e volute  of  the  curve  given  by  the  parametric  equations 
x=  a  (cos  6  +  0sin  0),  y  =  a  (sin  0  —  0cos  0). 

MISCELLANEOUS   EXERCISES 

Find  the  radius  of  curvature  and  coordinates  of  the  center  of  curvature 
of  the  following  curves  at  the  points  indicated. 

1.   y2  =  8x  +  l,  at  (6,  7).  2.    y  =  3 x3  -  8 x  +  4,  at  (-  2,  -  4). 

3.    y  =  cos  ac,  at  (0,  1).  4.   y  =  xex,  at  (0,  0). 

5.  Show  that  the  evolute  of  an  arch  of  the  cycloid  consists  of  the  halves 
of  an  equal  cycloid. 

6.  Show  that  the  radius  of  curvature  at  any  point  of  a  cycloid  is  double 
the  length  of  the  normal  at  the  same  point. 

7.  If  the  equation  of  a  curve  can  be  written  in  the  form 

y=  ±  (ax+  b)  +0(x), 

where  </>(#)  is  a  rational  fraction  the  denominator  of  which  is  higher  degree 
than  the  numerator,  show  that  the  lines  y  =  ±  (ax  +  b)  are  asymptotes  to 
the   curve.      In  this  way  determine  the  asymptotes   to  the  curve 
xs  —  xy2  +  ay'2  =  0. 

8.  Show  that  the  ordinates  of  the  curve  xs  =  xy  +  1  =  0  and  of  the 
parabola  y  =  x2  approach  the  same  value  as  x  increases.  For  this  reason,  the 
parabola  is  said  to  be  a  curvilinear  asymptote  to  the  curve. 

9.  Trace  the  curve  y  +  xy  —  xz  =  0.  Find  its  rectilinear  and  curvilinear 
asymptotes. 

10.  Show  that  at  the  point  (0,  0)  the  curve  y  =  — * —  has  two  terminat- 
ing branches  with  different  tangents.  ,    .    - 

Examine  for  singular  points  the  following  curves. 

11.  x(x  -  ay  +  y2(x  -  2  a)  =  0. 

12.  (x2  +  y2)*  =  4x2  +  y2. 

13.  x*  -  3  xy  +  6  xj2  -  y*  =  0. 

14.  By  the  method  of  limiting   intercepts,  find  the  asymptotes  to  the 

curves  (a)  y»  =  4  st*  -f  «8 ;  (b)   **  -  v-  -  1 . 

a'2     b2 

15.  Find  the  radius  of  curvature  of  the  three-cusped  hypocycloid  from 
the  parametric  equations 

x  =  o(2  cos  B  +  cos  2  0),   y  -  «(2  sin  6  -  sin  2  6). 


168  CURVES  [Chap.  IX. 

16.  The  tractrix  is  a  curve  having  the  property  that  the  length  of  the  tan- 
gent is  a  constant  a.     Find  the  equation  of  this  curve. 

17.  Show  that  a  curve  whose  equation  is  given  in  polar  coordinates  has 

dfi 
an  asymptote  when  the  subtangent  p2  —  is  finite  for  p  =  <x>  • 

dp 

18.  Using  the  result  of  Ex.  17,  find  the  asymptotes  of  the  following  curves  : 

(a)  p  =  a  tan  9  \       (b)  p  =  —  ;       (c)  p  =  a  sec  2  d. 

19.  Prove  that  the  evolute  of  the  logarithmic  spiral  p  =  aek0  is  a  similar 
logarithmic  spiral. 

20.  Find  the  equation  of  the  evolute  of  the  cissoid 

2  a  —  x 

21.  A  circle  of  radius  b  rolls  on  a  fixed  circle  of  radius  a,  a  >  6,  and  a 
point  on  the  rolling  circle  generates  an  epicycloid.  Show  (a)  that  the  equa- 
tions of  the  curve  are 

x  =  (a  +b)  cos  0  —  6  cos  g  "*"     <p, 
b 

y  =  (a  +  b)  sin  <p  —  b  sin  a  +     <p  ; 
b 

(6)  that  the  radius  of  curvature  is  4  h(a  +  ^  sin  ^- 

v  J  a  +  26  2b 

22.  Discuss  the  family  of  curves  obtained  by  giving  p  different  constant 
values  in  the  equation  c  =  a  +  /3r  +  p(l  +  -  p)- ,  which  gives  the  rela- 
tion between  specific  heat  c  and  temperature  Tof  superheated  steam. 

Xs 

23.  Trace  the   curve   y2  = Examine   for  maxima  and  minima, 

x—  a 

asymptotes,  points  of  inflexion,  and  cusps. 

7?  T        n 

24.  Van  der  Waals'  equation  p  =  ^= —  gives  the  relation  between 

v  —  b      v2 

pressure,  volume,  and  absolute  temperature  of  certain  substances.  Giving 
T  different  constant  values,  a  family  of  curves  called  isothermals  are  ob- 
tained. For  carbon  dioxide  B  =  0.00369,  a  =  0.00874,  b  =  0.0023.  Trace 
the  isothermals  for  T=  250,  300,  350. 


CHAPTER   X 

DEFINITE   INTEGRALS 

97.  Definition  of  a  definite  integral.  Suppose  we  have  given 
a  function  f(x)  which  is  continuous  and  single-valued  within  a 
given  interval  (a,  6),  and  suppose  further  that 

J>0)  dx = D^/ix) =<t>(x)+ a  "(l) 

The  difference  of  the  values  of  the  function  <f>  (x)  +  C  for  any  two 
values  of  the  independent  variable  is  called  a  definite  integral. 
The  values  of  the  independent  variable  substituted  are  called  the 
limits  of  integration.  We  denote  a  definite  integral  symbolically 
by  writing  the  two  limits  of  integration  at  the  extremities  of  the 
sign  of  integration.  Thus,  if  x  =  a  and  x  =  b  are  the  limits  of 
integration  in  the  definite  integral  formed  from  (1),  we  indicate 
that  fact  by  writing  this  definite  integral  as  follows : 


x 


f{x)  dx  =  lD^f(x)-]a  =  *  (6)  -  *  (a) .  (2) 

This  symbol  is  read :  "  The  definite  integral  of  f(x)  between  the 
limits  x  =  a  and  x=  b"  or  "from  a  to  6."  It  is  to  be  noted  that 
in  the  definite  integral  the  constant  of  integration  disappears. 
This  result  follows  from  the  definition ;  for,  we  have 

[>  (x)  +  0]\  =  [>  (b)  +  0]  -  [>  (a)  +  0]  =  <f>  (b)  -  *(a).       (3) 

To  distinguish  definite  integrals  from  those  previously  discussed, 
we  call  the  latter  indefinite  integrals.  We  may  not  only  pass  from 
the  indefinite  to  the  definite  integral  by  the  process  already  indi- 
cated, but  conversely  we  may  pass  from  the  definite  to  the  indefinite 
form  of  the  integral  by  assuming  the  upper  limit  of  integration 
as  variable.     Thus,  we  have 


s: 


/(»)(&  =  <£(»)-<£  (a), 


where  <£  (a)  may  be  taken  as  the  constant  of  integration. 

169 


170  DEFINITE    INTEGRALS  [Chap.  X. 


EXERCISES 

Verify  the  following : 

1.    £  4 a*  <te  =  624.  2.    C  (3  x  -  x8)  dx  =  f . 

sin0d6>=:  1 -cos/3.  4.     f   — =  *. 

>   VY-x2     « 

5.     ("»<?»=  log  «*•  6.     f4^  =  log4. 

-  5 

7.     f     tan<pd(p  =  0.  8.     f  2  sine  cos  Odd  =  £. 

Evaluate  the  following  definite  integrals  : 

Jl   3  +  X2  J2 

9  n 

11.    J     cos2  6de.  12.     Psinadte. 

7T 

13.     f^cosfldtf.  14.     C(—^-J\    dx,r<a. 

•'-f  J°  \a2  —  as2/ 

15.   ^S(a*-**)a?"*<to.  16.    f'sinfld*. 

17    rf  coBgdtf .  18   rVTT»^-x*dx. 

Jo  l  +  sin20  Jo 

19.  Show  from   known  formulas  of  physics  that  the   definite  integral 
\  2  gt  dt  gives  the  space  traversed  by  a  falling  body  in  the  time  interval  £2  —  h- 

20.  Plot    the    curve    v=f(t)=gt   and    show   from    geometry   that  the 
area   between    this   curve,    the    £-axis,    and   the  ordinates   at  t\  and  to  is 

numerically  equal  to  \  2  gt  dt. 

21.  Evaluate  f  (3  x2  —  4)  x  dx  and  f  (3  x2  —  4)x  dx.  What  is  the  re- 
lation between  the  two  integrals  ?  State  a  general  law  covering  the  inter- 
change of  the  limits  of  integration. 

22.  Show  that  V  {x2  -  3  x)  dx  +  £  (x2  -  3  x)  dx  =  C  (x2  -  3  x)  dx. 
What  general  law  does  this  result  suggest? 

23.  Show  that  the  general  theorems  for  indefinite  integrals  (Art.  67) 
apply  also  to  definite  integrals. 


Art.  98]  PROPERTIES   OF   DEFINITE   INTEGRALS  171 

98.  Elementary  properties  of  definite  integrals.  The  general 
properties  of  integrals  already  developed  apply  equally  well  to 
definite  integrals.  Thus,  the  definite  integral  of  a  constant  times 
a  function  is  equal  to  that  constant  times  the  definite  integral 
of  the  function ;  the  sum  of  a  finite  number  of  definite  integrals 
is  equal  to  the  definite  integral  of  the  sum  of  the  functions,  etc. 
In  addition  to  these,  there  are  certain  properties  which  apply  to 
definite  integrals  alone.  Among  the  more  elementary  of  these 
properties  are  the  following : 

Theorem  I.  Interchanging  the  limits  of  integration  changes  the 
sign  of  the  integral  ;  that  is, 

Xb  fa 

f(x)dx  =  -  I    f(x)dx. 

From  the  definition  of  a  definite  integral,  we  have 

f/(aO  <**  =  *(&)-*(<*),  (1) 

and  f'fW  dx  =  *  ^  *"  *® '  (2) 

Hence,  f  /(»)  dx  =  -  J  /(*)  dx.  (3) 

Theorem  II.     If  c  lies  between  a  and  b,  then 

fbf(x)  dx  =  fef(.&)  dv+   C  /(a?)  dap. 

•Ja  Ja  Jc 

For,  we  have 

£f(x)dx=4>(b)-f(a),  (4) 

£f(x)dx=cf>(c)-cf>(a),  (5) 

£f(x)dx=<j>(b)-<l>(c).  •       (6) 

Adding  (5)  and  (6),  we  obtain 

f /(a)  dx  +  Cf(x)  dx  =  c£(c)  -  <f>(a)+  $(b)  -  <^,(c) 

=jf  >(*)*»,  (7) 


172  DEFINITE   INTEGRALS  [Chap.  X. 

which  establishes  the  theorem.  Evidently  the  theorem  may  be 
extended  to  include  any  finite  number  of  values  between  a  and  b. 

EXERCISES 

1.  Evaluate  the  integral  I  (3  x2  +  5)  dx,  using  successively  the  limits  of 

integration  1  to  3,  3  to  4,  4  to  6,  and  1  to  6,  and  verify  Theorem  II  by  the 
result. 

2.  By  evaluating  the  integrals,  show  that 

IT 

ysinddd  +  ("sinddd  =  f  "sin  Odd. 

2 

3.  Show  that  i    0(w)  du  =  I    0(x)  dx.     State  this  result  in  the  form  of  a 

Ja  Ja 

theorem. 

99.  Change  of  limits.  In  the  process  of  finding  an  integral,  it 
is  sometimes  convenient  to  change  the  variable  (see  Art.  71).  In 
this  case,  we  may  by  properly  changing  the  limits  of  integration 
obtain  the  definite  integral  without  a  second  substitution.  Sup- 
pose we  have  the  integral   I    f(x)  dx  and  make  the   substitution 

z  =  F(x).  For  x  =  a,  we  have  z  =  F(a),  and  for  x  =  b,  z  =  F(b)\ 
hence  the  substitution  of  F(a)  and  F(b)  as  limits  of  integration  in 
the  transformed  integral  must  lead  to  the  same  result  as  the  sub- 
stitution of  a  and  b  in  the  original  integral. 


Ex.   Find    (*    arc  Sln  x  dx ,  assuming  z  -  arc  sin  x. 
Jo       VT^ 

We  have  dz  =      dx      .  whence  arc  sin  x  dx  =  z  dz. 

VI  -x2  Vl  -  x2 


For  x  =  0,  z= arc  sin  0  =  0,  the  lower  limit,  and  for  x  =  |,  z  =  arc  sin  \  =^  , 
the  upper  limit. 


Hence 


/*2  arc  sin  x  dx . 

=P 

»<**=£• 

J°      Vl  -  x2 

72 

EXERCISES 

r2  xdx 

Jo  1  4-  x* ' 

Let  z  =  x2. 

r2     <?x 

Let  a  =  - . 

Jo   Va2-x2 

a 

Arts.  99,  100]  DEFINITE   INTEGRAL   AS  A   SUM 


173 


3.  (     sin2  x  cos  x  dx. 

Jo 

4.  i   ex2xdx. 

5.  C      dx      . 

Jo  ex  -f  e~x 

77 

6     f  4  sin  fl  dd  < 

Jo     cos30 
?     f2  (log2  a)  da 


8. 


r 


dx 


(a2  +  x2)' 


Let  2;  =  sin  x. 
Let  2  =  x2. 
Let  0  =  ex. 

Let  ^  =  tan  0. 
Let  2  =  logx. 
Let  z  —  arc  tan 


AY 


EXP~ 


IOO.  Definite  integral  as  the  limit  of  a  sum.  In  Art.  7  it  was 
shown  that  the  area  underneath  a  curve  might  be  obtained  as  a 
limit  of  the  sum  of  certain  rectangular  elements.  As  pointed  out 
in  that  connection,  the  method  there  employed  is  impracticable 
and  except  in  the  most  elementary  cases,  perhaps,  impossible. 
We  shall  now  show 
the  relation  of  the 
summation  process 
to  the  definite  in- 
tegral; in  fact,  we 
shall  show  that  under 
certain  restrictions 
as  to  the  character  of 
the  function,  the 
limit  of  the  sum  of 
an  indefinitely  large 
number  of  indefin- 
itely small  elements  may  always  be  replaced  by  a  definite  integral 
taken  between  the  proper  limits. 

For  this  purpose,  let /(as)  be  a  single- valued,  continuous  function 
in  the  interval  a<?x<Lb,  and  suppose  it  be  represented  by  some 
curve  as  EF,  Fig.  51.  Let  the  interval  (a,  b)  be  divided  into  sub- 
intervals  by  the  insertion  in  any  manner  whatever  on  the  X-axis 
of  the  points  x1}  x2,  •  •  •,  xn_x  between  A  and  B.  Let  each  subinter- 
val  be  multiplied  by  the  value  of  f(x)  at  some  point  within  it. 


x=b 


+X 


Fig.  51. 


174  DEFINITE   INTEGRALS  [Chap.  X. 

Denoting  these  values  off(x)  by  f(x),  f2(x),  •••,/„(#)  respectively, 
we  have  then  the  sum 

2}  /<(»)*»<  =  (ft  -  «)  /i(»)  +  <>2  -  »r)  /2(»)  +  •••+(&-  ®»_i)  /„(«), 

(1) 
where  Aa^  denotes  any  one  of  the  subintervals  (xt  —  a^).     In  each 

subinterval  Ax{,  let  a{,  /3{  denote  the  smallest  and  largest  numerical 

values,  respectively,  of  f(x).     With  Axf  as  a  base  let  two  rectangles 

be  constructed  having  a{  and  ft  respectively  as  their  altitudes. 

Repeating  this  construction  for  each  subinterval,  we  have  the 

relation  »  «  » 

V  a^Aa,  ^  V  /^Aa;,  1  V  ftAav  (2) 

i  i  i 

The  first  and  the  last  of  these  sums  have  limits  as  n  increases 
indefinitely  since  both  are  monotone  functions  and  limited  in 
magnitude  by  the  conditions  placed  upon  fix).  Moreover  these 
limits  are  equal,  say  equal  to  A,  and  that  independently  of  the 
manner  of  subdividing  the  given  interval  (a,  &).*  Since  A#  =  0 
as  n  becomes  infinite,  we  may  now  write  (see  Art.  13) 

L    Vf(x)±x  =  A.  (3) 

The  existence  of  the  limit  in  (3)  being  established,  we  may  find 
the  value  of  the  limit  by  taking  the  subintervals  and  the  corre- 
sponding values  of  the  function  in  any  particular  manner  we 
please.  It  is  convenient  to  make  the  values  of  Asc  equal  and  to 
take  the  value  of  fix)  at  the  beginning  of  each  interval.  We 
have  then  for  all  values  of  n, 

nAx=b  —  a,  (4) 

and  the  original  points  of  division  become 

a  +  A#,     a  -h  2  Ax,  •  •  •,     b  —  Ace. 

In  defining  the  definite  integral,  <£  (a?)  was  taken  as  a  function 
having  f(x)  as  its  derivative.  From  the  law  of  the  mean,  we  have 
therefore 

<f>(x  +  Aa;)  -  <f>(x)  =  Ax  •  f{x  +  0  ■  Aa),     0  <  6  <  1.  (5) 


*See  Picard's  Traite  &  Analyse,  Vol.  1,  p.  4  ;  or  Veblen  and  Lennes'  In- 
finitesimal Analysis,  p.  150  et  seq. 


Art.  100]  THE   LIMIT   OF  A   SUM  175 

By  hypothesis  a  and  /3  are  respectively  the  minimum  and  maxi- 
mum values  of  f(x)  in  a  subinterval ;  hence  we  may  write 

a^f(x  +  6-  Ax)^/3. 

By  aid  of  (5)  this  inequality  becomes 

a  •  Ax  ^  <£>  (<e  +  Aa;)-</.(a;)^i8.  Ax.  (6) 

We  may  now  apply  (6)  to  the  successive  subintervals  by  sub- 
stituting for  x  the  values 

a,     a  +  Ax,     a  +  2  Ax,  •  •  •,     b  —  Ax. 

The  resulting  inequalities  are : 

«!  Ax  ^  <£  (a  +  Ax)  —  <£  (a)  <J  /?!  Ax, 

«2  Ax  <^  <£ _(a  +  2  Ax)  —  <£  (a  -f  Ax)  <^  /?2  Ax, 


an  Ax  ^  <£  (6)  -  <£  (6  -  Ax)  ^  ft,  Ax. 
Adding  these  results,  we  have 

6  6 

2}  «,Ax^  <£(&)  -  £(a)  ^  2}  ft  Ax.  (7) 

a  a 

Since  we  have 

a  a 

it  follows  that  <f>(b)-cf>  (a)  =  A,  (9) 


and  we  may  therefore  write 

b  r* 

AJlo]£/0*0A«  =  <i>(&)-4>(a)=  I    f(x)dx.  (10) 


We  may  henceforth  replace  at  any  time  the  summation  pro- 
cess   by   the    definite   integral,    and    regard    the    two    symbols, 

A     .0X     an(^      I        as  interchangeable  whenever  the  integrand 

a 

is  a  continuous  function. 

From  this  discussion  it  also  follows  that  the  definite  integral 


176  DEFINITE   INTEGRALS  [Chap.  X. 


»6 


f   fix)  dx  is  represented  graphically  by  the  area  bounded  by  the 
curve  y  =f(x),  the  X-axis,  and  the  two  ordinates  x  =  a,  x  =  b. 

lOl.  Importance  of  the  summation  process.  The  result  just 
derived  and  expressed  by  equation  (10)  of  the  preceding  article  is 
of  the  highest  importance,  for  it  enables  us  to  replace  a  difficult  and 
tedious  process  of  direct  summation  by  a  process  that  is  iti  most 
cases  simple  and  easily  carried  out.  As  was  shown  in  the  example 
of  Art.  7,  the  determination  of  the  area  under  a  curve  involves 
the  summation  of  an  indefinitely  large  number  of  indefinitely 

small    terms,   that    is,   it    requires   the   limit        ^_  Q  V  f(x)  Ax. 

a 

But  according  to  (10)  the  limit  is  given  by  the  definite  integral 
j  f(x)  dx.  Hence  to  find  the  area  we  have  only  to  find  the  anti- 
derivative  <£  (x)  of  the  given  function  fix),  substitute  the  limits  a 
and  b,  and  take  the  difference  <f>  (5)  —  <f>  (a).  For  example,  to  find 
the  area  under  the  curve  y  =  x2  from  the  origin  to  the  ordinate 
x  =  3  (see  Art.  7),  we  have 

A=  Cf(x)dx=  Cx*dx  =  \x*f  =  §. 

While  the  discussion  leading  to  equation  (10)  was  accompanied  by 
a  geometrical  illustration,  and  the  summation  was  directed  toward 
the  determination  of  an  area,  the  course  of  reasoning  depends  in 
no  way  upon  geometrical  considerations.  The  method  of  the 
definite  integral  is  applied  with  equal  facility  to  the  determination 
of  magnitudes  of  all  kinds  —  volumes,  masses,  fluid  pressures, 
heat,  work,  etc.  In  the  following  chapters  there  will  be 
given  examples  of  the  use  of  the  summation  process  in  finding 
the  lengths  of  curves,  the  areas  of  the  surfaces,  the  volumes  of 
solids,  etc.  In  mechanics,  determinations  of  centers  of  gravity 
and  moments  of  inertia  likewise  involve  the  summation  principle. 
The  work  done  by  a  variable  force  is  found  by  the  summation  of 
terms  of  the  type  FAs,  where  ^denotes  force  and  s  displacement. 
The  impulse  of  a  variable  force  is  the  summation  of  terms  of  the 
type  F&t.  The  space  over  which  a  moving  point  travels  is  found 
by  the  summation  of  terms  of  the  type  vAt,  where  v  denotes  the 


Art.  101]       IMPORTANCE    OF   THE    SUMMATION   PROCESS        177 

velocity  of  the  point.  If  the  specific  heat  c  of  a  substance  varies 
with  the  temperature  r,  the  heat  that  must  be  imparted  to  the 
substance  to  produce  a  given  rise  of  temperature  is  determined 
by  the  summation  2c  At.  Other  applications  will  occur  to  the 
student. 

It  should  be  noted  that  the  summation  of  an  infinite  number  of 
terms  is  always  necessary  when  one  of  the  factors  entering  into 
the  problem  varies  continuously.  As  an  illustration,  take  the 
problem  of  finding  the  mass  of  a  body.  If  V  denotes  the  volume 
of  the  body  and  y  its  density,  the  product  yV  gives  the  mass, 
provided  the  density  is  constant  throughout.  The  density,  how- 
ever, may  differ  for  different  parts  of  the  body,  as,  for  example, 
when  the  body  is  composed  of  different  liquids  which  arrange 
themselves  in  layers  or  strata.  If  Vlt  V2,  Vs,  •••,  Vn  denote  the 
volumes  of  the  separate  parts,  and  ylf  y2,  y3,  •••,  yn  the  correspond- 
ing densities,  the  total  mass  is  evidently  the  sum 

yiFi  + y^  +  ^-f  •-+7,K. 
In  this  case,  the  number  of  parts  being  finite,  we  need  only  simple 
addition.  However,  the  density  may  vary  continuously  through- 
out the  body,  as  in  the  case  of  the  atmosphere.  Here  we  must 
have  recourse  to  the  summation  of  an  infinite  number  of  indefi- 
nitely small  terms.  We  divide  the  total  volume  V  into  n 
parts  each  equal  to  AF  and  multiply  each  element  AV  by  the 
density  at  that  part  of  the  body.  We  thus  get  n  terms  of  the 
type  7  A  V.  If  n  is  finite,  the  sum  of  these  n  terms  is  not  the  exact 
value  of  the  mass  because  the  density  varies  in  the  element  of 
volume  A  V.  But  as  n  is  taken  larger  and  A  V  correspondingly 
smaller,  the  sum  of  the  n  terms  approaches  more  nearly  the  mass. 
Hence,  to  get  the  exact  result,  we  must  increase  n  indefinitely,  thus 
making  AF  correspondingly  small,  and  effect  the  summation  of  the 
infinitely  large  number  of  infinitesimal  terms.     That  is,  we  must 

find  L      T7AF. 

As  previously  shown,  the  summation  is  effected  most  easily  by 
means  of  the  definite  integral.  The  elements  to  be  summed  being 
of  the  type  f(x)  Ax,  we  find  the  anti-derivative  <j>(x)  of  the  function 
/(#),  substitute  the  limits  of  integration,  say  a  and  b,  and  take 


178 


DEFINITE   INTEGRALS 


[Chap.  X. 


Fig.  52. 


the  difference  <£(&)  —  <f>(a).     The  problem  of  effecting  the  sum- 
mation reduces,  therefore,  to  a  problem  in  integration. 

Ex.     A  vertical  wall  (as  a  dam)  having  a  height  h  and  breadth  6,  Fig.  52, 
is  subjected  to  water  pressure,  the  intensity  of  which  varies  as  the  depth 

below   the   liquid  surface.      Required  the 
total  pressure  on  the  wall. 

According  to  the  law  of  liquid  pressure, 
the  intensity  at  the  depth  x  is  kx,  where  k 
is  constant.  Let  the  wall  be  divided  into 
elements  of  width  Ax  and  length  b ;  then 
if  the  area  of  the  strip  b  Ax  is  multiplied 
by  the  intensity  of  pressure  kx  at  the  top  of 
the  strip,  the  product  b  Ax  •  kx=kb  x  Ax  gives 
approximately  the  pressure  on  the  element  of  area.  The  sum  of  a  finite  num- 
ber of  terms  of  the  type  kbxAx  would  give  a  total  pressure  somewhat  smaller 

h 

than  the  actual  value;   but  the  limit      L     2]kbxAx  evidently  gives  the 

exact  result.     This  limit  is  the  definite  integral  (    kbx  dx  =  kb  \    xdx  = 
\  kbh2.     Hence  the  total  pressure  on  the  wall  is  \  kbh2. 

102.  Geometrical  representation  of  a  definite  integral.  It  was 
pointed  out  in  Art.  100  that  a  definite  integral  I  f(x)  dx  is  repre- 
sented by  the  area  bounded  by  the  curve  y  =f(x)f  the  X-axis,  and 
the  ordinates  corresponding  to  x  =  a,  x  =  b.  Whatever  magni- 
tude the  definite  integral  is  used  to  denote,  volume,  mass,  fluid 
pressure,  work,  or  moment,  this  area  is  the  graphical  representa- 
tion of  it;  that  is,  the  number 
of  units  of  area  is  the  same  as 
the  number  of  units  of  the  mag- 
nitude denoted  by  the  integral. 
In  fact,  one  way  of  evaluating  a 
definite  integral  is  to  measure  by 
some  mechanical  means  the  area  that  represents  it.  (See  Art.  119.) 

If  f(x)  becomes  negative  for  certain  values  of  x,  the  graph  of 
fix)  will  lie  below  the  X-axis,  as  from  b  to  c,  Fig.  53.     In  this 

case  the  integral   J    fix)  dx  is  represented  by  the  algebraic  sum 
of  the  areas  I,  II,  and  III,  area  II  being  taken  as  negative.     If 


Fig.  53. 


Arts.  102,  103]     GEOMETRICAL   REPRESENTATION  179 

the  numerical  rather  than  the  algebraic  sum  is  desired,  we  must 

Xb  f*c 

fix)  dx,    I    f(x)  dx,  and 
Jb 

fix)  dx  without  reference  to  sign. 

Ex.  Show  that  the  area  which  represents  the  total  liquid  pressure  in 
the  example  of  the  preceding  article  is  a  triangle  whose  base  is  h  and  whose 
altitude  is  kbh.     (Note  that/(x)  =  kbx.) 

EXERCISES 

1.  Give  a  geometric  interpretation  of  Theorem  II,  Art.  98. 

2.  Give  a  geometric  proof  of  the  following  theorem  :  If  M  and  N  are 
respectively  the  greatest  and  least  values  of  the  continuous  function  fix) 
within  the  interval  (a,  b),  then,  provided  b  >  a, 

•5 


N(b  -  a)<  f  f(x)  dx  <  M(b  -  a). 


3.  Give  a  geometric  proof  of  the  following  theorem  :  If  b  >  a,  and  <f>(x), 
f(x),  and  \f/(x)  are  three  functions  such  that  for  any  value  of  x  within  the 
interval  (a,  6),  <p(x)  <f(x)  <$ix),  then 

f  (f>(x)dx<  Cf(x)dx<  C\p(x)dx. 

4.  Using  the  theorem  of  Ex.  3,  show  that  if  0  <  n  <  2,  (  —  -  lies 
between  0.5  and  0.7854.  J°  1  +  x" 

5.  Give  a  geometric  proof  of  the  theorem 

C  F(x)  dx  =  (*°  F(a  -  x)  dx. 

6.  Show  geometrically  that  f    sin2  6  dd  -  \ "  cos2  d  dd. 

7.  Show  from  geometric  considerations  that 

i      sin  0  d0  =  0. 

8.  Show  the  area  that  represents  the  definite  integral  (  2  gt  dt,  which 
applies  to  falling  bodies.  n 

103.  Definite  integrals  of  discontinuous  functions  :  Infinite  limits 
of  integration.  So  far  we  have  discussed  definite  integrals  of  con- 
tinuous functions  with  finite  limits  of  integration.  We  shall 
now  consider  cases  where  one  or  both  of  these  conditions  do  not 
hold. 


180 


DEFINITE   INTEGRALS 


[Chap.  X. 


Suppose  that  f(x)  has  a  point  of  discontinuity  at  x  =  c,  Fig.  54. 
The  extent  of   this  discontinuity  may  be  finite  or  infinite.     In 

fix)  dx,    I    f(x)  dx,    I    fix)  dx  have 

no  meaning  according  to  the  defi- 
nition given  for  a  definite  integral. 
However,  the  definite  integral  does 
have  a  meaning  for  the  intervals 
(a,  c  —  e),  (c  +  e,  b) ;  for  within 
these  intervals  the  function  is 
continuous.      We  may,    therefore, 

define  the  value  of  the  definite  integral  for  the  intervals  (a,  c) 

and  (c,  b)  as  the  limits 

L    (      f(x)  dx,  L    I    f(x)  dx, 

provided  these  limits  exist.  The  definite  integral  for  the  inter- 
val (a,  b)  may  now  be  defined  as  the  sum  of  these  two  limits. 
We  may  not  always,  however,  obtain  the  proper  value  of  the 

last  integral  by  evaluating  directly  the  integral    I   f(x)  dx. 

1 


Ex.  1.     Find  the  area  between  the  curve  y  =  — 

(x 
the  ordinates  for  which  x  =  0,  x  =  4.  v 


0,  x 

This  function  is  discontinuous  for  x 
nite  integral  for  the  given  interval  is 
found  by  taking  the  sum  of  the  limits 


2)2 


the  X-axis,  and 


2,  as  is  shown  in  Fig.  55.     The  deri- 


Hence  the  area  in  question  is  infinite. 

Let  us  now  take  the  integral  directly  between  the  limits  0  and  4.     We 
obtain  for  the  area 


Rw-K  «£&-;=&' 


i, 


Art.  103]  INFINITE   LIMITS   OF   INTEGRATION  181 

an  incorrect  result.     In  this  case  we  say  therefore  that  the  integral  1  f(x)  dx 
has  no  meaning. 

We  must  also  consider  the  special  case  in  which  one  of  the 
limits  of  integration  is  infinite.  This  again  is  a  limiting  case  of 
the  ordinary  definite  integral,  and  we  define  the  definite  integral 

fix)  dx  as  the  limit    L     J    f(x)  dx.      Hence,    provided    this 

limit  exists,  the  integral    I    fix)  dx  has  a  meaning.     The  follow- 
ing example  illustrates  this  case : 

Ex.  2.  Find  the  area  between  the  X-axis  and  the  witch  of  Agnesi,  whose 
equation  is 

y=     8a3 


xl  +  4  a1 

See  Fig.  39,  Art.  90.     The  curve  is  symmetrical  with  respect  to  the  T-axis 
and  approaches  the  X-axis  as  x  becomes  infinite.     Hence  the  area  is  given  by 

2(*yclx=    L    I6a3p— ^— -  =  16  a8    L    P-  arc  tan  -^-1 
Jo   *  9smao         Jo  x2  +  4  a2  a-=ooL2a  2aJo 

=  8  a2   L   arc  tan  —  =  4  ira2. 

X=ao  2  d 

EXERCISES 

1.    Evaluate   f"—^— .  2.    Evaluate   f   e~axdx. 

Jo  a2  +  x*  J° 

3.    Evaluate  C ^ 4.    Evaluate   f'1 — -^ 

Jo  (1-*)*  xVx*=l 

5.    Of  the  following  definite  integrals,  which  have  finite  values  ? 


«/■*;  <»  ff ;  (0  JTS;  <«  ft 


(?.'• 


.'•- 


6.  Show  that  if  ft  <  1 ,  the  area  under  the  curve  xny  =  C  from  the  origin 
to  x  =  a  is  finite,  while  for  n  >  1  the  area  under  the  curve  from  x  =  a  to 
x  =  oo  is  finite. 

7.  With  the  data  of  Ex.  6  investigate  the  area  when  n  =  1. 

8.  Find  the  area  under  the  curve 

2/(z2  -  1)*  =  X 

between  x  =  0  and  x  =  4.     Draw  the  curve. 


182  DEFINITE   INTEGRALS  Chap.  X. 

MISCELLANEOUS   EXERCISES 

Evaluate  the  following  definite  integrals  : 

l.    ["V^db.  2.   ^a(xVa^^  +  ^^\dx. 

\ 

•2 


4.   Show  that  ^r  =  o«« 


3 .     f  2  sin2  6  cos3  0  dd.     (Put  cos2  6  =  1  -  sin2  6. ) 
(x(a  —  x)dx 
i    (a  —  «)  da; 

( ^  sin  0  cos  Odd              \    acos20sm0d0 
5.    Evaluate  (a)  Jft  _fl  — ;    (&)  ^- • 


f  ^sinfl 


cos  0  sin  0  cZ0 


6.  Find  by  the  summation  method  of  Art.  7  the  area  between  the  curve 
y  =  ex,  the  X-axis,  and  the  ordinates  x  =  1,  x  =  4. 

7.  Change  the  limits  of  integration  of  the  following  integrals  when  the 

variable  is  changed  as  indicated. 

/•20  ri 

(a)    \     (  )dx,  x  =  z2  -j-4;  (6)    I    (  )dx,  x  =  sin0  —  cos0; 

/•log  10  /•« 

(c)    j         (  )  dx,  ex  =  22  +  1  ;      (d)    I    (  )dx,  x  =  asin  9. 

8.    Show  from  geometrical  considerations  that 

(«)    (*5c/(X)dx  =  c£f(x)dx  ;    (6)    Cf(x)dx  =C~af(x  +  a)  eta. 

Evaluate  the  following : 

9.    f     a*      .  io.     C      dx       .  11.     P— *5 

J°  V2^  Jo  Va2  -  x2  Jo  O  -  2)2 

12.  Prove  and  interpret  geometrically  the  following  theorem  : 

(a)  »/(-«)  =  -/(*),  |V(x)dx  =  0. 

(6)  If/(-x)=/(x),    Cf(x)dx  =  2$oaf(x)dx. 

13.  Apply  the  theorems  of  Ex.  12  to  the  definite  integrals 


J     sin  6  dd  ;    f     cos  0  c?0. 
7T  ./       7T 


_7T 
2  2 


Art.  103]  MISCELLANEOUS  EXERCISES  183 

14.  If  n  >  2,  show  that  f  *      dx       lies  between  0.5  and  0.52. 

Jo    VI -xM 
fg 

15.  If  /(x)  =  V8  x  +  10,  show  that  I   /(x)  dx  must  lie  between  3  V50  and 

3V74. 

16.  Find  the  limits  between  which  the  integral  f  /  (x)  dx  must  lie  where 
/  (x)  =  togas. 

17.  Prove  the  following  theorem  geometrically : 

f  0(x)  dx  =  (6  -  a)0[a  +  0(6  -  o)], 
where  0  <  d  <  1. 


CHAPTER   XI 


APPLICATIONS    OF   INTEGRATION   TO  GEOMETRY   AND 
MECHANICS 

104.    Plane    areas,  rectangular   coordinates.      If    the    equation 

of    a    curve    in    rectangular    coordinates    is    y  —f(x),  then,    as 

seen  in  Art.   100,  the   area   A   between  the  curve,  the  X-axis, 

and  the  ordinates  x  =  a  and  x=b,  respectively,  is  given  by  the 

formula  /•&  rb      ,  ... 

^4=  J    /(as)  das  =  J    ydvc.  (1) 

Ex.  1.   Find  the  area  included  between  the  parabola  y2  =  8  x,  the  X-axis, 
the  origin,  and  the  ordinate  x  =  18. 
Since  y2  =  8  x,  y  =  V8  x,  and 

.4  =  f 18 w  dx  =  V8  f 18 x* dx  =  V8[| x*]18  =  144. 
Jo  Jo  ° 

Ex.  2.   Find  the  area  under  one  arch  of  the  curve  y  =  cosx. 


Fig.  56. 


From  the  graph  of  the  curve,  Fig.  56,  it  is  seen  that  y  =  0  for  x  =  • 

for  x  =  -  •     Hence,  we  have 
2 

7T  7T 

r2  i2 

area  .45(7=  I     cosxdx  =  sinx        =2. 

"2  ~2 

3ff  3tt 

Likewise,  area  QBE  =  f     cos  x  dx  =  sin  x  1      =  -  2. 


and 


184 


Art.  104]       PLANE  AREAS,  RECTANGULAR  COORDINATES      185 


Also, 


area  ABODE  —  sin  x 


0,  and  area  BCD 


=  sin  x 
J( 


The  last  two  results  are  consistent  when  the  signs  of  the  areas  are  taken  into 
account.  Thus  areas  ABC  and  CDE  are  equal  in  magnitude  but  opposite  in 
sign.     The  numerical  magnitude  of  the  area  ABCDE  is  4. 

It  will  be  observed  that  the  area  ABC  is  symmetrical  with 
respect  to  the  axis  OF;  consequently 


area  ABC  =2  x  area  OBC=2  C 
Likewise,     area  CDE  =  2  J     cos  x  dx  =  —  2. 


cos  x  dx  =  2. 


In  general,  it  is  advisable  to  make  use  of  conditions  of  symmetry 
as  far  as  possible,  and  take  the  narrowest  interval  of  integration 
that  the  problem  permits. 

Frequently  it  is  desirable  to  find  the  area  between  the  curve, 
the  Faxis,  and  the  abscissas  corresponding  to  y=a  and  y  =  fi, 
respectively.     The  formula  for  this  area  is 


=  f  ocdy. 


(2) 


The  derivation  of  form  (2)  follows  precisely  that  of  form  (1 )  and 
need  not  be  given  in  detail. 

Ex.  3.  Find  the  area  between  the  curve 
ys  =  kx,  the  F-axis,  and  the  abscissas  for  y  —  2 
and  y  =  3. 


*-J>*-sJ>*-i-M 


65 
4k' 


>X. 


Ex.  4.  Find  the  area  included  between  the 
hyperbola x?/=36 and  the  straight  linex+i/  =  15 
(Fig.  57). 

The  points  of  intersection  of  the  two  curves 
are  found  by  solving  the  equations   simultaneously.     The    solution   gives 
(3,  12)  and  (12,  3)  as  the  coordinates  of  A  and  B,  respectively. 

Area  ABCD  =  area  EACBF -  area  EADBF 

'dx 
x 

r2 


/M2  /»12. 

j     (15-a0<fa-36l    - 
15  x  ---36  log x 


17.59. 


186  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

105.  Plane  areas,  polar  coordinates.  When  polar  coordinates 
are  used,  the  area  swept  over  by  the  radius  vector  in  passing 
from  an  initial  position  to  a  final  position  is  the  required  area. 

Let  AB,  Fig.  58,  represent  the  curve  p=f($),  where  f(0)  is  a 
continuous  function.     Let  OX  be  the  initial  line,  and  a  and  {3  the 


Fig.  58. 

initial  and  final  values  of  6.     The  area  required  is  that  of  the 
sector  AOB. 

Let  the  angle  AOB  =  j3  —  a  be  divided  into  n  parts,  A6lf  A02, 
etc.  In  the  ^th  division,  let  pj  and  pt"  be  respectively  the  smallest 
and  largest  values  of  p.  If  from  0  as  a  center,  arcs  are  drawn 
with  p-  and  p"  as  radii,  two  circular  sectors  are  formed  whose 
areas  are  respectively  i  p/2  A0<  and  |p/'2A^-.  Proceeding  in  this 
way  with  each  of  the  n  divisions,  we  get  n  circular  sectors  with 
arcs  lying  within  the  given  curve  AB,  and  n  circular  sectors  with 
arcs  lying  without  the  curve.  The  required  area  OAB  is  greater 
than  the  sum  of  the  inner  sectors  and  less  than  the  sum  of  the 
outer  sectors  ;  that  is, 

^}lP'2A0<2iTea,AOB<^^-lp"2A0.  (1) 

a  a 

8  8 

Since  ^hp^AO  and   V{/)"2A^  always  remain  finite,  the  first 


Art.  105]  PLANE   AREAS,  POLAR   COORDINATES  187 

never  decreasing  and  the  second  never  increasing,  each  sum  has  a 
limit,  and  by  the  method  used  in  Art.  100,  it  may  be  shown  that 
these  limits  are  equal.     We  have,  therefore, 

area  AOB=     L     Ti/2A^=     L    Y|y,2A<9.  (2) 

Furthermore,  since  p  is  a  continuous  function  of  6,  we  may 
replace  the  common  limit  by  a  definite  integral  and  write 

-4=|£VcW,  (3) 

where  A  denotes  the  required  area. 

Ex.  1.    Find  the  area  swept  over  by  the  radius  vector  of  the  curve  p  =  - , 
as  6  varies  from  ir  to  2  it. 

a?  C2lT  dd  _  a? 
4 


2Jn  2  J*      02 


Ex.  2.    Find  the  area  swept  over  by  the  radius  vector  of  the  logarithmic 
spiral  p  =  eke  in  one  revolution. 


2>  4  k       Jo        4kK  J 


EXERCISES 

1.  Find  the  area  between  the  line  y  =  3  x  +  4,  the  X-axis,  and  the  lines 

x  =  o,  x  =  a. 

2.  Find  the  area  between  the  parabola  y  —  5  x2,  the  X-axis,  and  the  lines 
x  =  0,  x  =  4. 

3.  Find  the  area  under  the  curve  5  y2  =  x3  from  the  origin  to  the  line  x=5. 

4.  Find  the  area  between  the  parabola  y=Sx2  and  the  straight  line  y  =  12  x. 

5.  Take  the  arc  of  the  equilateral  hyperbola  xy  =  C  between  the 
points  A  and  B.  Show  that  the  area  between  this  arc  and  the  X-axis  is  the 
same  as  the  area  between  the  same  arc  and  the  F-axis. 

X  X 

6.  Find  the  area  under  the  catenary  y  =  -  (ea+  e  B)  between  x  =—  m 
and  x  =  m. 

7.  Derive  a  general  expression  for  the  area  under  the  curve  xym  =  C 
between  x  =  a  and  x  =  b.     Discuss  the  case  m  =  1 . 

8.  Find  the  area  swept  over  by  the  radius  vector  of  the  spiral  of  Archi- 
medes p  =  ad  in  one  revolution. 


188 


APPLICATIONS   OF   INTEGRATION 


[Chap.  XL 


9.    Find  the  area  of  the  two  loops  of  the  lemniscate  p2  =  a2  cos  2  0.     Make 
use  of  the  symmetry  of  the  curve  and  take  narrowest  limits  of  integration. 

10.  Find  the  area  swept  over  in  two  revolutions  by  the  radius  vector  of 
the  parabolic  spiral  p2  =  a2d. 

11.  Find  the  total  area  bounded  by  the  curve  a*y2  +  b2x*  =  a2b2x2. 
Suggestion  :  Trace  the  curve  in  order  to  obtain  proper  limits  of  integration. 

12.  Find  the  area  between  the  curve  y2(a2—x2)  =  a2x2,  the  asymptote 
x  =  a,  and  the  X-axis. 

13.  Show  that  the  area  bounded  by  the  spiral  pd  =  C  and  two  radii  pi  and 
P2  is  proportional  to  p\  —  pi- 

14.  Find  the  area  of  a  loop  of  the  curve  p2  —  a2  sin  nd. 

106.  Volumes  of  solids  of  revolution.  A  plane  area  AEFB, 
Fig.  59,  bounded  by  the  curve  EF,  whose  equation  is  y  ==/(*), 
the  axis  of  X,  and  the  ordinates  AE  and  BF,  rotates  about  the 

axis  OX  and  thereby 
generates  a  solid  of 
revolution.  To  de- 
rive an  expression 
for  the  volume  of 
this  solid  we  pro- 
ceed as  follows :  Let 
the  interval  AB  be 
divided  into  n  parts, 


fT 


ffp^ 


+X 


Fig.  59. 


x=b 

and  let  ordinates  be 

erected  at  the  points 
of  division.  In  any  subinterval  Axt,  let  ?//  and  yf"  be  respectively 
the  smallest  and  largest  numerical  values  of  the  ordinate,  and 
construct  rectangles  having  Axt  as  a  base  and  y-  and  y-\  respec- 
tively, as  altitudes.  Repeating  this  construction  for  each  of  the 
subinterval s,  we  obtain  one  plane  area  made  up  of  rectangles 
lying  entirely  below  the  given  curve  EF  and  a  second  plane 
area  made  up  of  rectangles  whose  upper  bases  lie  above  this 
curve.  The  solids  obtained  by  revolving  these  areas  about  the 
X-axis  have  respectively  the  volumes 

F'  =  ]T  tt^Az,     V"  =2}  Try"2  Ax. 


Art.  106]  VOLUMES   OF   SOLIDS   OF   REVOLUTION  189 

The  volume  V  of  the  solid  generated  by  the  revolution  of  the 
area  AEFB  must  lie  between  V  and  V".     That  is, 

b  b 

]T  Try'2  Ax  ^  V^  ]P  Try"2  Ax.  (1) 

a  a 

It  will  be  seen  that  V  and  V"  are  both  monotone  functions,  the 
first  never  decreasing  and  the  second  never  increasing ;  hence 
since  the  functions  are  finite,  each  has  a  limit  as  Ax  =  0  (Art.  14). 
Following  the  method  of  Art.  100,  it  may  be  shown  that  the  two 
limits  are  equal.     Consequently  we  may  write 

b  b 

V=L    Viry^Ax^L    V  rry"2  Ax.  (2) 

Since  y  is  taken  to  be  a  single-valued  and  continuous  function  of 
x,  we  may  replace  the  common  limit  by  the  definite  integral  and 
write 

V=ir£y^dx.  (3) 

By  a  similar  process  it  can  be  shown  that  the  volume  of  the 
solid  generated  by  a  rotation  about  the  F-axis  is 

V=-*§*^dy9  (4) 

where  c  and  d  are  the  ordinates  of  the  end  points  of  the  curve. 

Ex.  1.  Find  the  volume  generated  by  the  revolution  about  the  X-axis  of 
the  area  bounded  by  the  line  4  x  +  y  =  12  and  the  coordinate  axes. 

For  y  =  0,  x  =  3  ;  hence  the  limits  of  integration  are  x  =  0,  and  x  =  3. 

V=w  Cy*dx  =  TT  f3  (12-  4x)2dx  =  7r(144x-48x2  +^V)]3  =  144  r. 
Jo  jo  3        Jo 

Ex.  2.  Find  the  volume  generated  by  the  rotation  about  the  A"-axis  of 
the  area  bounded  by  the  segment  of  the  parabola  y2  =  8  x  between  the  origin, 
the  X-axis,  and  the  ordinate  for  x  =  6. 


V  =  ir  Cifdx  =  ir  f 6  8  x  dx  =  4  7tx2~|    =144 

about  the  F-axis  of  the  area  bounded  by  th< 
cissa,  and  the  same  segment  generates  the  vo 

V  =  irCTS*2dy=*CrHy-<ly=  ^T4T  =  ^(48)t  =  156.72. 
Jo  y         J        04  y       320 Jo  320 V     ; 


The  rotation  about  the  F-axis  of  the  area  bounded  by  the  F-axis,  the  cor- 
responding abscissa,  and  the  same  segment  generates  the  volume 


190  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

EXERCISES 

1.  A  point  (to,  n)  is  joined  to  the  origin  by  a  straight  line.  Find  by 
integration  (a)  the  volume  of  the  cone  generated  by  revolving  the  line  about 
the  axis  OX;  (&)  about  the  axis  OY. 

2.  Find  the  volume  of  the  solid  generated  by  the  revolution  about  the 
X-axis  of  a  segment  of  the  equilateral  hyperbola  xy  =  12  between  x  =  4  and 
x  =  16. 

3.  Find  the  volume  generated  by  the  same  segment  about  the  Y-axis. 

4.  The  segment  of  the  curve  y  =  sec  x  between  x  =  0  and  x  =  \  t  is  re- 
volved about  the  X-axis.     Find  the  volume  of  the  solid  generated. 

5.  Find  the  volume  generated  by  the  revolution  of  the  catenary 

about  the  X-axis,  taking  x  =—  c  and  x  =  c  as  limits. 

6.  Find  the  volume  generated  by  revolving  about  the  X-axis  the  part  of 
the  parabola  x%  +  y^  —a^  intercepted  by  the  axes. 

7.  Find  the  volume  of  the  solid  generated  by  revolving  about  the  X-axis 
the  plane  area  bounded  by   the  X-axis,   the  line  x  =  a,  and  the  cissoid 

y*  =      x*      • 
y       2a-x 

8.  The  area  lying  to  the  left  of  the  X-axis,  between  the  exponential 
curve  y  =  ex  and  the  X-axis,  is  revolved  about  the  X-axis.  Find  the  volume 
of  the  solid  thus  generated. 

107.  Volumes  determined  by  the  summation  of  thin  slices.  The 
method  of  determining  a  volume  by  taking  the  limit  of  a  number 
of  thin  slices,  which  we  have  used  in  finding  the  volume  of  revo- 
lution, can  be  extended  to  other  solids.  We  conceive  the  solid  to 
be  cut  by  a  number  of  parallel  planes  usually  perpendicular  to  one 
of  the  coordinate  axes.  In  this  way,  the  solid  is  divided  into  a 
number  of  thin  slices  lying  along  the  axis  in  question.  Suppose 
the  cutting  planes  to  be  chosen  perpendicular  to  the  X-axis ;  the 
thickness  of  the  slice  may  be  denoted  by  Ax,  and  the  area  of  the 

cross  section  will   be   a  function  of    x,    say    F(x).     Then   the 

h  b 

volume  Flies  between  two  sums  "F' =^«  Aa;  and  V"  =^)(3 &x, 

a  a 

where  -a   and   (3   denote   respectively   the   smallest   and   largest 


Art.  107] 


VOLUMES   OF   SOLIDS 


191 


values    of    F(x)   within  the  subinterval   Ax ;    and,  furthermore, 
these  sums  approach  the  same  limit  as  Ax  =  0.     Hence  we  have 

V=    L   y.F(x)Ax=  f  F(x)dx.     [Art.  100.]  (1) 

AX  =  0    a  Ja 

The  limits  of  integration  must  be  so  chosen  as  to  include  all  the 
slices. 

It  will  be  observed  that  (1)  has  the  same  form  as  the  formula 
for  a  plane  area.  The  Fix)  which  gives  the  area  of  the  cross  sec- 
tion is,  however,  generally  different  from  the  f(x)  which  gives  the 
ordinate  of  the  bounding  curve. 

Ex.  1.  A  circular  cylinder  of  length  h  is  cut  by  a  plane  which  passes 
through  the  diameter  of  one  base  and  is  tangent  to  the  other  base.  Required 
the  volume  of  the  piece  cut 
from  the  cylinder. 

Taking  the  axes  as  shown 
in  Fig.  60,  let  the  solid  be 
divided  into  slices  by  planes 
parallel  to  the  ]TZ-plane.  The 
sections  cut  by  the  planes  are 
evidently  similar  triangles. 
For    any    triangle    as    ABC, 

AB  ='y,  and  BC  =  ^-h,  where 
a 

a  is  the  radius  of   the   base ; 

hence  the  area  is 


2a 


(a2-x2)h 
2a 


Fig.  60. 


For  the  total  volume  we  have  therefore 
h 


V  = 


2a 


jo> 


x*2)  dx  —  -  alh. 
J  3 


Ex.  2.    Find  the  volume  of  the  ellipsoid  —  +  v—  +  —  =  1. 

n%        h'2        r% 


a*     b*     & 

Let  the  cutting  planes  be  taken  parallel  to  the  TZ-plane  ;  then  a  plane 
section  at  a  distance  x  from  the   FZ-plane  will  be  an  ellipse  having  for  its 


equation 


b*      c2 


This  equation  may  be  written 


&2 


r 

(a2  -  x2) 


(a2-x2) 


192  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

The  semiaxes  of  the  ellipse  are  therefore  -  Va2  —  x2  and  -  Va2  —  x2,  and  the 

7  .  #  Q> 

area  of  the  ellipse  is  —  {a2-  x2). 
a2 
Hence,  we  have 


a2    Jo  3 


rabc. 


EXERCISES 

1.  Find  the  volume  of  the  elliptic  paraboloid  ^-+  —  =  4  x  between  the 
planes  x  =  0  and  x  =  3. 

2.  Find  the  volume  of  the  solid  bounded  by  the  surface  —  +  2-  +  —  =  1. 

tf      b2     c2 

3.  By  the  method  of  integration  derive  the  formula  for  the  volume  of  a 
pyramid  or  cone. 

4.  Derive  the  formula  for  the  volume  of  the  frustum  of  a  pyramid  or 
cone. 

5.  A  right  circular  cylinder  of  base  radius  a  and  altitude  h  has  two  slices 
cut  from  it  by  planes  passing  through  a  diameter  of  one  base  and  touching 
the  other  base.     Find  the  volume  of  the  wedge-shaped  solid  remaining. 

6.  A  cap  for  a  post  is  a  solid  of  which  every  horizontal  cross  section 
is  a  square,  and  the  corners  of  the  squares  lie  in  the  surface  of  a  sphere 
14  inches  in  diameter  with  its  center  in  the  upper  face  of  the  cap.  The  depth 
of  the  cap  is  4  inches.     Find  the  volume. 

7.  A  solid  is  formed  by  a  variable  square  whose  center  moves  along  the 

major  axis  of  the  ellipse  —  +  *-  =  1  in  such  a  way  that  the  side  of  the  square 
25       9 

is  always  equal  to  the  double  ordinate  of  the  ellipse.     Find  the  volume  of  the 

solid. 

108.  Lengths  of  curves,  rectangular  coordinates.  The  length  of 
a  curve  may  be  defined  as  the  limit  of  the  sum  of  the  lengths  of 
the  inscribed  chords  as  the  number  of  the  chords  is  increased 
without  limit  and  the  length  of  each  approaches  the  limit  zero. 

Given  the  curve  AB,  Fig.  61,  whose  equation  is  y=f(x),  where 
f(x)  is  a  continuous  function  having  a  continuous  derivative. 
The  length  s  of  this  curve  between  the  limits  x  =  a,  x  =  b  is 
required.  Divide  the  given  interval  b  —  a,  into  n  equal  parts, 
denoting  by  Ax  one  of  these  equal  divisions.  Denote  by  Ac, 
Ay,   respectively,   the   corresponding  values   of    the    chord   and 


Art.  108] 


LENGTHS   OF   CURVES 


193 


of  the  increment  of  y.     From  the  definition  of  the  length  of  a 
curve,  we  may  now  write 


i±»~*i,*FJE** 


AX  =  0    a 


Az  =  0 


W 


(i) 


With  the  restrictions  imposed  upon  y=f(x),  the  quotient  ^,  by 

Ax 


*  X 


Fig.  61. 


dy 


the  law  of  the  mean,  is  equal  to  the  derivative  —  for  some  value 

dx 

of  x  within  the  subinterval  Ax.     Hence,  since  by  Art.  100  we 
may  take  the  value  of   the  integrand  \1  +  (  —  )    at   any  point 

within  this  subinterval,  we  may  in  (1)  replace  —  by  —  and  write 

Ax        dx 


ii^RSV 


Replacing  the  summation  by  the  definite  integral,  we  have  finally 


■■£$+( 


doc/ 


dx. 


(2) 


If  y  is  taken  as  the  independent  variable,  this  formula  for  the 
length  of  a  curve  becomes 


L'lMgr* 


(3) 


where  c,  d   are  the  values  of  y  corresponding   to   x  =  a,  x  =  6, 
respectively. 


194  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

Ex.  1.   Find  the  length  of  the  curve  y3  =  3  x2  from  the  point  (0,  0)  to  the 
point  (3,  3) . 

From  the  given  equation  we  have 

dx  _  y2 

dy     2x' 

•-JCV1 +(!)"* = 1  £  V¥T^dy = I (4 + 8  *>t|! 

=  J r  (13^  -4$)  =4.32. 

Ex.  2.    Find  the  length  of  the  cycloid  described  by  a  point  on  the  circum- 
ference of  a  rolling  disk  during  one  revolution  of  the  disk. 
The  equations  of  the  cycloid  are 

x  =  a(0  —  sin0), 
y  =  «(1  —  cos0), 
in  which  0  denotes  the  angle  through  which  the  disk  has  turned.      Differen- 
tiating,  we  have  <fe  =  «(1  -  cos  *)  <**, 

dy  =  a  sin  0  dd, 
whence  dx2  +  dy2  =  2  a2(\  —  cos  0)  dd2. 

From  (2),  we  may  write  s  =  l  y/dx2  +  dy2, 

whence  s  =  ( (dx2  +  dy2)^  =  aV2  1(1  —  cos  0)^  dd. 

Now  Vl  —  cos  0  =  V2  sin  \  0,  and  the  limits  of  0  are  obviously  0  and  2  tr  for 
one  revolution. 

Hence  s  =  2a\     sin-  d0  =  —  4  a  cosr  |     =  8  a. 


2j0 


EXERCISES 


1.  Find  the  length  of  the  curve  x1  +  y'  =  a7. 

2.  Find  the  length  of  the  semicubical  parabola  ay2  =  xs  from  the  origin  to 
the  point  (m,  n). 

3.  Find  the  length  of  the  catenary  y  =  \  a (e«  +  e~<0 ,  from  x  =  0  to  x  =  set* 

4.  Find  the  circumference  of  a  circle  (a)  using  the  equation  x2  +  y2—  a2; 
(b)  using  the  equations  x  =  a  cos  0,  y  =  a  sin  0. 

5.  The  involute  of  a  circle  is  given  by  the  equations 

x  =  a(cos  0+0  sin  0) , 
y  =  «(sin  0—0  cos  0). 
Find  the  length  of  an  arc  between  the  limits  0  =  0  and  0  =  w. 


-JN 


Art.  109]  LENGTHS   OF   CURVES  195 

109.  Lengths  of  curves,  polar  coordinates.  The  formulas  of  the 
preceding  article  for  finding  the  length  of  a  curve  in  rectangular 
coordinates  may  be  so  transformed  as  to  cover  the  case  of  polar 
coordinates  by  means  of  the  equations 

x  =  p  cos  6,        y  —  p  sin  6.  (1) 

From  Art.  108,  we  have 

r^gv  (2) 

or  introducing  the  differential  dx  under  the  radical, 

s  =  CVda?  +  dy*.  (3) 

From  equation  (1)   we  obtain 

dx  —  —  p  sin  0d0-\-  cos  0  dp  1  , . 

dy  =  p  cos  0  dQ  +  sin  0  dp     J  ' 

Substituting  these  expressions  in  (3),  we  have 

Veto2  +  dy2  =  V[(—  p  sin  6  dd  +  cos  0  dp)2  +  {p  cos  OdB  +  sin  8  dp)2] 

=  ^P2d0i  +  dp2.  (5) 

Consequently,  we  have 

where  a,  /?  are  the  limits  of  integration  corresponding  to  the  limits 
a,  b  when  written  in  rectangular  coordinates. 

When  p  is  taken  as  the  independent  variable,  this  formula  may 

be  written  ,.*    I         Jd»\*.  m 

Ex.     Find  the  whole  length  of  the  cardioid 

p  =  2a(l  —  cos0). 
Differentiating  the  equation  of  the  curve,  we  get 

^  =  2  a  sin  0. 
dd 

Substituting  in  (6),  we  have 


2  C  n2  a[(l  -  cos  0)2  +  sin2  0]*  dd 

Xir  6  r  &ln 

4asin-<Z0  =  16a    —  cos-      =  16  a. 


196  APPLICATIONS   OF  INTEGRATION  [Chap.  XL 

EXERCISES 

1.  Find  the  length  of  the  circumference  of  the  circle  p  —  2  a  cos  0. 

2.  Find  the  length  of  the  logarithmic  spiral  p  =  eae  from  6  =  0  to  6  =  —  - 

also  from  6  =  -  to  6  =  v. 
2 

3.  Find  the  length  of  the  spiral  p  =  euB  from  the  pole  (p  =  0)  to  p  =  1. 

4.  Find  the  length  of  an  arc  of  the  conchoid  p  =  a  sec  6  between  0  =  0  and 


HO.  Areas  of  surfaces  of  revolution.  Let  a  plane  curve  y  =f(x) 
revolve  about  the  X-axis  and  thereby  generate  a  surface.  We 
may  find  an  expression  for  the  area  8  of  this  surface  by  a  method 
similar  to  that  employed  in  finding  the  length  of  the  curve. 

Divide  the  interval  AB,  Fig.  61,  into  n  equal  parts,  and  denote 
the  length  of  each  by  Ax.  Denote  by  Ac  the  length  of  the  corre- 
sponding chord  PQ,  by  Ay  the  corresponding  increment  of  y,  and 
by  AS'  the  surface  generated  by  revolving  the  chord  Ac  about  the 
axis  of  X  Since  the  lateral  area  of  the  frustum  of  a  cone  of 
revolution  is  the  product  of  the  slant  height  and  one  half  the  sum 
of  the  circumferences  of  the  bases,  we  have  for  the  element  of 
surface  thus  generated 

AS'  =  2Jy±^\cA.  (1) 

But  Ac  =  VA^+A/ =  \  1  +  (^Y  Ax ;  (2) 

hence,  taking  the  sum  of  all  the  elements  of  surface  thus  formed, 
we  have 

2>S'  =  22„(,±f)VlT(gjA*.  (3) 

As  Ax  =  0,  the  left-hand  member  of  this  equation  approaches  the 
area  of  the  surface  generated  by  the  revolution  of  the  given  curve, 
that  is,  the  required  area.  In  fact,  we  may  define  the  area  of 
the  required  surface  of  revolution  as  the  limit  of  the  sum  of  the 
surfaces  of  these  frustums  as  Aa?  =  0.     In  the  risrht-hand  member 


r±f 


j  approaches  y,  and  a  /l  -f-  (  — )  becomes  \1  +[  —  ]  >  since 


Art.  110]        AREAS   OF   SURFACES   OF   REVOLUTION  197 

y  —f{x)  is  assumed  to  be  a  continuous  function  having  a  continu- 
ous derivative.     We  have  then 

or  S  =  2  7rCbyyjl  +fS$doB.     (Arts.  100,  108)         (5) 

When   OF  is  taken  as   the  axis  of   revolution,  the  formula 
becomes 


'-■•r-Vi+K)'*"-       <«> 


From  (2)  it  is  evident  that  we  may,  if  we  choose,  replace 


V 


^'ti" 


in  (5)  and  (6)  by  ^l+ffjdy.    ' 

In  some  problems  the  latter  form  is  more  convenient.  The  limits 
of  integration  must  be  changed  to  the  values  of  y  corresponding 
to  x  =  a,  x  =  b. 

Ex.  1.      Find    the    surface    generated    by    revolving    the    hypocycloid 

%*  +  y*  =  a1  about  the  .X-axis. 
From  the  given  equation,  we  get 


We  have  therefore 


\dy)       \y) 


■i-J>g)**-^.S*»I 


!*•?**!  =j™*- 


Ex.  2.     Find  the  area  of  the  surface  generated  by  the.  revolution  of  a 
cycloid  about  its  base. 

The  equations  of  the  cycloid  are 

x  =  a(6  -  sin0),  y  =  a(l  —  cos0). 
Differentiating  these  equations,  we  have 

dx  =  a(l  —  cos  0)  dd, 
dy  —  a  sin  6  dd, 

whence         ^i  +  /<& V dx  =  VdWTW2  =  aV2(l  -  cosd)dd. 


198  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

Substituting  in  (5),  we  get 


S  -  2  Tra2  Cw  V2(l  -  cos  ey  dd 


-• 'JT-g)-®-! 


—  TO? 

3 


EXERCISES 

1.  Find  the  area  of  the  surface  generated  by  the  revolution  of  the  parabola 
y2  =  8  x  about  the  X-axis.  Take  the  limits  x  =  0  and  x  =  2  ;  also  the  limits 
x  =  2  and  x  =  8. 

2.  A  line  joins  the  origin  to  the  point  (w,  ri).  Find  by  integration  the 
surface  of  the  cone  generated  by  revolving  this  line  about  the  X-axis. 

3.  Find  the  area  generated  by  revolving  about  the  A'- axis  the  arc  of  the 
cubical  parabola  2  y  =  x8  between  x  =  0  and  x  =  2.' 

4.  Find  the  area  of  the  surface  generated  by  revolving  about  the  X-axis  the 

X  X 

arc  of  the  catenary  y  =  -  (ea  +  e  a)  between  x  =  0  and  x  =  a. 

z 

111.  Mean  value.  Let  y  =/(a?)  be  a  continuous  function  within 
the  interval  x  —  a  and  x  =  b,  and  suppose  this  interval  to  be 
divided  into  n  equal  parts  each  equal  to  Aa,\  Then  b  —  a  =  n  Ax. 
Denoting  by  ylf  y2,  y3,  •••,  yn  the  values  of  the  function  corre- 
sponding to  the  values  of  x  at  the  middle  points  of  these  successive 
subdivisions,  let  us  form  the  quotient 

yi  +  ^  +  y3+  •••  +y» 

n 

which  evidently  is  merely  the  arithmetic  mean  of  the  n  values  of 
?/.  This  quotient  will  vary  with  the  number  of  divisions  n,  and 
the  limit  which  it  approaches  as  n  is  indefinitely  increased  is 
called  the  mean  value  of  the  function  for  the  interval  b  —  a. 

Substituting  for  n  its  value     ~a ' ,  (1)  may  be  written 

Ax 

yx  Aa;  +  y2  AaH h  y„  Ax 

b  —  a 
The  limiting  value  of  this  quotient  as  n  becomes  infinite,  and  con- 
sequently as  Ax  =  0,  is 

b-  a  b  —  a  W 


(1) 


Art.  Ill] 


MEAN   VALUE 


199 


The  numerator  of  (2)  is  represented  by  the  area  under  the  curve 
x=f(x)  (as  AEFB,  Fig.  62)  between  the  ordinates  for  x  =  a  and 
x  =  b.  Hence  if  a  rectangle 
AMNB  is  constructed  having 
an  area  equal  "to  the  area  under 
the  curve,  the  altitude  of  this 
rectangle  represents  the  quotient 


i 


ydx 


Fig.  62. 


b-a   ' 
or  the  mean  value  of  the  func- 
tion. 

The  independent  variable  may 
be  time,  distance,  angle,  area, 
volume,  or  any  other  geometrical  or  physical  magnitude.  Mean 
values  may  be  taken  with  reference  to  different  variables.  Thus 
in  the  case  of  a  moving  point,  values  of  the  velocity  may  be 
taken  for  equal  time  intervals  or  for  equal  space  intervals.  In 
the  former  case,  the  mean  velocity  will  be  the  mean  ordinate 
of  the  velocity  curve  on  a  time  base ;  in  the  latter  case,  it  will  be 
the  mean  ordinate  of  the  velocity  curve  on  a  space  base. 

Ex.  1.     In  simple  harmonic  motion  the  velocity  of  the  moving  point  is 

v  =  a  sin  wt, 

and  the  time  of  a  half-oscillation  is  t  =  ir/w. 

Hence,  the  mean  velocity  for  the  time  interval  0  to  ir/ut  is 


a  \    sin  cot  dt 


T     a  ,T 

COS  u>t 

L      o  J( 


--0 


Ex.  2.    In  the  case  of  a  falling  body  we  have 


v  =  gt 


and 


v2  =  2  gs. 

For  the  mean  velocity,  taking  time  as  the  variable,  we  have  for  the  inter- 
val 0  to  «!, 

j>«   ^l'tdt   i 


=  2ffh- 


200  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

Since  the  velocity  at  the  end  of  the  time  h  is  gh,  the  mean  velocity  is  one 
half  the  final  velocity. 

The  mean  velocity  for  the  space  Si,  taking  equal  space  intervals,  is 

(  lvds      V'2g  I  lVsds      0     0 

si  —  0  si  3  d 

that  is,  the  mean  velocity  is  two  thirds  the  final  velocity. 


EXERCISES 

1.  Find  the  mean  of  the  ordinates  of  the  parabola  y2  =  10  x  from  x  =  0  to 
x  =  8. 

2.  Find  the  mean  ordinate  of  the  curve  y  =  x2  —  7  x  +  5  between  x  =  1 
and  x  =  5. 

3.  Find  the  mean  value  of  the  ordinates  of  the  curve  y  =  cos  x  (a)  between 
x  =  0  and  x  =  |  it  ;  (&)  between  x  =  0  and  x  =  7r. 

4.  A  number  n  is  divided  into  two  parts  ;  find  the  mean  value  of  the  prod- 
uct of  the  parts. 

5.  During  the  expansion  of  steam  in  an  engine  cylinder  the  pressure  falls 
approximately  according  to  the  law  px  =  C,  where  x  denotes  the  distance  the 
piston  has  moved  from  the  beginning  of  the  stroke.  Derive  an  expression 
for  the  mean  pressure  exerted  during  the  period  of  expansion. 

6.  Find  the  mean  ordinate  of  a  semicircle  of  radius  a  provided  the  ordi- 
nates are  drawn  at  equal  intervals  on  the  arc. 

Suggestion  :  Use  polar  coordinates. 

112.  Work  of  a  variable  force.  When  the  point  of  application 
of  a  force  is  moved  in  the  direction  of  the  line  of  action  of  the 
force,  the  force  is  said  to  do  work.  For  example,  work  is  done  by 
the,  drawbar  pull  of  a  locomotive  when  the  locomotive  moves 
along  the  track,  thus  moving  the  point  of  application. 

If  the  force  is  constant  in  magnitude,  the  work  done  is  the  prod- 
uct of  the  force  and  the  distance  through  which  the  point  of  appli- 
cation moves  in  the  direction  of  the  force.  If  IT  denotes  the  work, 
F  the  force,  and  s  the  displacement  of  the  point  of  application, 

then  W  =  Fs.  (1) 

For  example,  the  work  done  in  raising  a  load  of  800  pounds  a  height 
of  6  feet  is  800  X  6  =  4800  foot-pounds. 


Art.  112] 


WORK   OF   A   VARIABLE   FORCE 


201 


The  magnitude  of  the  force  may  vary  as  the  point  of  applica- 
tion moves,  as  in  compressing  a  spring.  In  this  case,  the  work 
may  be  expressed  as  a  definite  integral  as  follows :  Let  sx  and  s2 
be  the  initial  and  final  distances  of  the  point  of  application  of 
the  force  from  some  origin,  so  that  s2  —  sx  is  the  displacement. 
Let  s2  —  sx  be  divided  into  n  subintervals  Asx,  As2,  etc.,  and  in  any 
subinterval  A^  let  F-  and  F"  be  respectively  the  smallest  and 
largest  values  of  the  force  F.  Then  denoting  the  work  by  W, 
we  have 

^  F'  As  ^  W^  2)  F"  As-  (2) 

*2 

As  in  Art.  100,  we  may  show  that  the  functions  2  F'As  and 

2j  F"  As  have  the  same  limit  as  As  =  0,  and  we  may  therefore 
write  at  once,  since  F  is  a  continuous  function  of  s, 


W=    L 

As  =  0 


s^=r 


Fds. 


(3) 


From  the  method  of  deriving  (3),  it  is  clear  that  the  work  of 
a  force  may  be  represented  by  an  area.  On  the  displacement  SXS2 
let  the  forces  F  corresponding  to  the  successive  positions  of  the 
point  of  application  be  laid  off  as  ordi- 
nates,  and  let  a  curve  be  drawn  through 
the  ends  of  these  ordinates  (see  Fig.  63)  ; 
then  the  area  ^ABSi  under  this  curve 
will  represent  the  work,  of  the  force  be- 
tween Si  and  Sf.  For  the  area  under 
the  curve  is 

"  Fds, 


F 


F* 


Sj 


s 

Fig.  63. 


8m 


which  according  to  (3)  gives  the  work  W. 

In  order  to  integrate  the  expression  Fds,  the  force  F  must  be 
expressed  as  a  function  of  the  displacement  s.  The  following 
cases  are  those  occurring  most  frequently  in  practice. 


202  APPLICATIONS    OF   INTEGRATION  [Chap.  XI. 

(a)    When  the  force  is  a  linear  function  of  the  displacement. 
This  is  the  law  in  the  compression  of  a  spring.     We  have 

W  =  p  Qcs  +  b)ds  =  l  Jcs2  +  bs]*2 

=  lfc(822-Sl2)+K^2-S1). 

Ex.  1.   A  spring  is  10  inches  long,  and  a  force  of  48  pounds  is  required  for 

each   inch   it  is  compressed.     Find 


IV 


-10- 


the  work  of  compressing  the  spring 
— —      from    10    inches    to     a    length    of 
6  inches ;    also  from  a  length  of  8 
inches  to  a  length  of  5  inches. 
Fig.  64.  In  the  first  case, 

8l  =  0,  s2  =  10  -  6  =  4,  F  =  48  s. 

Hence,  W  =  f  *2  F  ds  =  48  f *  s  ds  =  24  iP~Y  =  384  in.  lb. 

For  the  second  case,  si  =  2,  s-2  =  5,  and 

W  =  48  C  s  ds  =  24  s2T  =  504  in.  lb. 

Ex.  2.  A  bar  is  stretched  from  its  original  length  L  by  a  gradually  in- 
creasing load.  Denoting  by  s  the  amount  of  stretch  for  a  given  force  F, 
Hooke's  law  gives  as  the  relation  between  F,  s,  and  L, 

w_EAs 

where  E  denotes  the  coefficient  of  elasticity  of  the  material,  and  A  the  area 
of  the  bar.     The  work  of  stretching  the  bar  by  an  amount  s  is  therefore 

Jo  L    Jo  2L 

We  have  for  wrought  iron  E  —  30,000,000  ;  suppose  a  bar  having  a  cross- 
section  area  of  2  square  inches  be  stretched  from  60  inches  to  60.5  inches. 
Here  ^4  =  2,  a  =  0.5,  and  L  =  60 ;  hence  the  work  is 

30000000  x  2  x  0.52  =         m  in  ft 
2  x  60 
(b)    When  the  force  varies  inversely  as  the  displacement. 
In  this  case,  & 

8 

whence  W=  fVjbsfc  f*2-  =fc  log-2- 


Arts.  112,  113]         WORK  OF   EXPANDING   GASES 


203 


(c)  When  the  force  varies  inversely  as  the  square  of  the  distance. 
The  law  of  the  inverse  square  applies  to  gravitational  forces,  to 
forces  between  electric  charges,  etc. 

Ex.  Let  a  positive  charge  m  of  electricity  be  concentrated  at  a  point  P, 
and  a  unit  charge  at  a  point  Q  at  a  distance  s  from  P ;  then  the  repulsion  of 
charge  m  on  the  unit  charge  is 

p  —    m    —  ™. 


PQ 


The  work  required  to  move  the  unit  charge  from  .«?  =  a  to  s  =  b  is 

Ja  Ja    S2  Sja  \a        b) 

113.  Work  of  expanding  gases.  A  gas  is  confined  by  the  walls 
of  a  cylinder  and  a  movable  piston,  Fig.  65.  By  virtue  of  its 
pressure,  the  gas  expands  or  increases 
in  volume,  moves  the  piston,  and  thus 
does  work  against  an  external  re- 
sistance. Let  p  denote  the  pressure 
exerted  by  the  gas  on  a  unit  area 
(square  inch  or  square  foot),  and  A 
the  area  of  the  piston.  Evidently  the  total  force  acting  on  the 
piston  is  F  =pA ;  and  for  F1  As,  F2  As,  etc.,  we  may  write  pxA  As, 
p2A  As,  etc.  But  A  As  is  the  volume  swept  over  by  the  piston 
in  moving  through  the  distance  As  and  may  be  denoted  by  Av; 
hence  F  As=p  Av,  and  the  work  done  is 


Fig.  65. 


v2  rv°- 

W=     L     V  p  Av  =  I    p 

Av  =  0  ^f  Jvi 


dv. 


In  using  this  formula  it  must  be  noted  (1)  that  p  denotes  pres- 
sure per  unit  area,  not  the  total  pressure,  and  (2)  that  for  correct 
numerical  results  consistent  units  must  be  used,  pounds  and  feet, 
or  pounds  and  inches  throughout;  thus,  if  p  is  in  pounds  per 
square  inch,  v  must  be  in  cubic  inches,  and  the  result  will  be  work 
in  incTi-pounds. 

Ex.  1.  Air  expands  without  change  of  temperature  following  Boyle's  law. 
pv  =p\Vi  =  const.     The  work  of  expansion  is 


W  =  Pp  dv  =  plVl  C2  *°  =  p1vl  log  * 

Jvx  Jvx     V  Vx 


w 


204  APPLICATIONS   OF   INTEGRATION  [Chap.  XI. 

Ex.  2.   Air  expanding  adiabatically  follows  the  law 
pvk  =  i>iVi*  =  const.,  where  k  =  1.41. 
The  work  done  during  the  expansion  is 

=  \  2pdv  =pxVik  \    «-*  dv  =  piVik— 

J"!  Ji  1  —  kJvt 

~*-iL  J~    /fc-i    ' 

since  Pi»i*  =  #*«**• 

EXERCISES 

1.  The  length  of  an  unstretched  spring  is  16  inches,  and  a  force  of  225 
pounds  is  required  to  stretch  it  1  inch.  Find  the  work  required  to  stretch  it 
from  a  length  of  18  inches  to  a  length  of  22.5  inches. 

2.  In  hoisting  coal  or  ore  from  a  mine,  the  load  consists  of  two  parts  : 
(1)  the  weight  M  of  the  car  and  contents  ;  (2)  the  weight  of  the  rope,  which 
is  m  pounds  per  foot.  Find  the  work  required  for  hoisting  a  distance  of  h 
feet. 

Suggestion  :  Let  s  denote  the  distance  of  the  load  from  the  lower  level ; 
then  F=M+m(h  —  8),  and 

W=  (  [M  +  m(h -  *)]  ds. 

3.  In  Exs.  1  and  2  draw  diagrams  showing  by  areas  the  work  done,  and 
derive  the  results  by  elementary  geometry. 

4.  Suppose  the  force  to  vary  directly  as  the  square  of  the  displacement  of 
its  point  of  application.     Derive  a  formula  for  the  work. 

5.  Confined  air  having  a  volume  of  6  cubic  feet  and  a  pressure  of  80  pounds 
per  square  inch  expands  following  the  law  pv  =  const,  to  a  final  volume  of 
20  cubic  feet.     Find  the  work  done. 

6.  Use  the  data  of  Ex.  5  and  find  the  work  done  if  the  expansion  is 
adiabatic,  i.e.  according  to  the  lawpu1-41  =  const. 

7.  Find  the  work  of  stretching  a  round  iron  bar  having  a  diameter  of 
1.5  inches  from  a  length  of  40  inches  to  a  length  of  42.3  inches.  Take 
E  =  28,000,000. 

MISCELLANEOUS   EXERCISES 

1.   Find  the  areas  bounded  by  the  following  curves,  the  X-axis,  and  the 
ordinates  indicated : 

(a)  y  =  x8  +  x  +  5,  from  x  =  0  to  x  =  6. 
(6)  y  =  xs  -  3  x2  +  4,  from  x  =  1  to  x  =  5. 
(c)  y  =  ex,  from  x  =  0  to  x  =  1. 


Akt.  113] 


MISCELLANEOUS   EXERCISES 


205 


2.  Find  the  area  of  one  loop  of  the  curve  p  =  a  sin  2  0. 

3.  Find  the  area  between  the  curve  y  =  tan  x  and  the  X-axis  from  x  =  0 


to*=-. 

3 

4.  Find  the  area  between  the  parabola  x^  +  y^  =  cfi  and  the  coordinate 
axes. 

5.  Find  the  volume  generated  by  the  revolution  of  the  entire  curve 
x2  +  y*=  1,  (a)  about  the  X-axis;  (&)  about  the  F-axis. 

6.  A  cylindrical  vessel  having  an  altitude  of  12  inches  and  a  base  diameter 
of  8  inches  is  tipped  and  the  contained  fluid  is  poured  out  until  the  liquid  sur- 
face coincides  with  a  diameter  of  the  base.  Find  the  quantity  of  liquid  re- 
maining in  the  vessel. 

7.  If  e  is  the  eccentricity  of  an  ellipse  and  <p  the  eccentric  angle,  the  para- 
metric equations  of  the  ellipse  are  :  x  =  a  cos  0,  y  =  b  sin  0.  Show  that  the 
entire  length  of  the  ellipse  is 


-j^ 


e2  cos2  cp  dp. 


8.  The  solid  shown  in  Fig.  66  is  gener- 
ated by  moving  a  variable  rectangle  DEFG 
parallel  to  the  plane  XOY.  One  angle  D 
moves  along  the  axis  OZ,  and  the  other 
angles  E  and  G  move  in  given  curves  on  the 
planes  TZ  and  ZX,  respectively.  If  the 
curves  QEB  and  PGB  are  circular  arcs  of 
radius  a  with  centers  at  0,  find  the  volume 
of  the  solid. 

9.  In  Ex.  8  find  the  convex  surface 
QSB. 

10.  In  Fig.  66,  take  OB  =  S,  OP=  0Q=6, 
and  assume  the  curves  BQ  and  BP  to  be 
parabolic  arcs  with  vertices  at  B. 
areas  of  convex  surfaces. 


Find  the  volume  of  the  solid  and  the 


11.  In  Fig.  66,  let  the  curves  BP  and  BQ  be  elliptic  quadrants  with  major 
and  minor  semi-axes  of  m  and  n  respectively.  Find  the  volume  of  the 
solid. 

12.  The  value  of  a  harmonic  alternating  current  is  given  by  the  equation 
i  =  i0  sin  0,  where  i0  is  the  maximum  value.  Find  the  mean  value  of  the 
current  for  a  half  cycle,  that  is,  from  0  =  0  to  6  =  ir. 


206 


APPLICATIONS   OF   INTEGRATION 


[Chap.  XI. 


13.    In  Fig.  67  is  shown  a  cylindrical  journal  and  bearing.     The  intensity 
of  the  normal  pressure  at  any  point  P  is  assumed  to  be  proportional  to  the 

depth  of  P  below  the  diameter  AG.  If 
Po  is  the  intensity  at  the  lowest  point  B, 
find  the  mean  intensity  over  the  surface 
ABC. 

14.  Find  an  expression  for  the  work 
done  by  a  gas  expanding  isothermally 
according  to  van  der  Waals'  equation 


~^> 

a! 

0 

m 

/ 

^t 

mm. 

B 

HP 

^ 

ft 

Fig. 

57. 

K)<°- 


b)=C 


from    an    initial   volume  v\  to   a  final 
volume  v2. 

15.  Find  an  expression  for  the  work 
done  by  a  force  that  varies  as  the  nth 
power  of  the  displacement  of  its  point  of 
application. 

16.  Find  the  work  required  to  compress  20  cubic  feet  of  air  from  a  pres- 
sure of  14.5  pounds  per  square  inch  to  a  pressure  of  63.5  pounds  .per  square 
inch,  the  equation  of  the  compression  curve  being  jw1,3  =  const. 

17.  A  particle  of  mass  m  has  simple  harmonic  motion  defined  by  the 
equation  x  =  r  cos  wt.  Show  that  the  mean  kinetic  energy  (£  mv2)  for  a  com- 
plete oscillation  is  one  half  the  maximum  kinetic  energy. 


CHAPTER   XII 
SPECIAL   METHODS   OF   INTEGRATION 

114.  Integration  by  parts.  Thus  far  we  have  made  use  of  only 
those  integrals  that  could  be  evaluated  by  use  of  the  fundamental 
formulas  given  in  Chapter  VI.  In  some  cases  we  were  able  to 
reduce  the  given  function  to  a  fundamental  form  by  a  simple 
transformation.  Not  all  functions,  however,  can  be  easily  re- 
duced to  those  types  by  the  methods  already  employed,  and  in 
this  chapter  we  shall  consider  some  special  methods  by  which 
this  reduction  may  be  effected  in  cases  more  complicated  than 
those  already  discussed. 

If  u  and  v  are  two  functions  of  the  same  independent  variable, 
we  have  upon  differentiating  their  product, 

d  (uv)  =  udv  -\-v  du, 

or  udv  —  d  (uv)  —  v  du. 

Integrating,  we  get 

I  udv  —  uv  —  \v du.  (1) 

By  this  formula  the  integral   J  u  dv  is  obtained  by  the  evaluation 

of  another  integral  J  v  du.     This  method  is  called  integration  by 

parts,  and  is  one  of  the  most  useful  of  the  integral  calculus.  It  is 
particularly  helpful  in  the  integration  of  the  product  of  two  func- 
tions where  the  integral  of  one  can  be  easily  found ;  also  in  the 
integration  of  logarithmic  functions,  exponential  functions,  and 
inverse  trigonometric  functions.  No  general  directions  can  be 
given  as  to  which  of  the  two  functions  is  to  be  taken  as  u  and 
which  as  dv,  except  that  the  selection  should  be  such  as  will 
render  dv  and  v  du  most  easily  integrable.     In  case  of  doubt,  first 

207 


208  SPECIAL   METHODS   OF   INTEGRATION       [Chap.  XII. 

try  putting  dv  equal  to  the  most  complicated  factor  that  can  be 
easily  reduced  to  a  fundamental  form. 

Ex.  1.     |  xex  dx. 

Put  dv  =  ex  dx,  u  —  x, 

whence  v  =  ex,  du  =  dx. 

Substituting  these  values  in  formula  (1),  we  have 


(  xex  dx  —  xex—  \  ex  dx  =  ex  (x  -  1)  + 


a 


Ex.  2.     (  x  arc  tan  x  dx. 

Put  dv  =  xdx,  u  =  arc  tan  x, 

whence  v  =  -  x2,  du 


2  1  +  x'2 

Substituting  these  values  in  the  formula,  we  have 

/•  i  \  C    x2 

\  x  arc  tan  x  dx  =  -  x2  arc  tan  x  —  \ dx 

J  2  2  J  1  +  x2 

=  -x2  arc  tan  x  -  -  ( dx  +  -  f — * 

2  2J  2J1  + 


dx_ 

x2 


—  -x2  arc  tan  x x +  -  arc  tan  x  +  0 

2  2         2 

=  a^  +  1  arc  tan  x  -  -  x  +  C. 

2  2 


,,| 


Ex.  3.    \  Va2  -  x2  dx. 


Put  u  =  V  a2  —  x2,        dv  =  dx, 

whence  du  = xdx     ,      v  =  x. 

Va2  -  x2 

The  result  of  substituting  these  values  in  (1)  is 

(Va2  -  x2  dx  =  xVa2^2  +  f-g^g — 
J  J  Va2  -  x2 

We  may  write 

*       =_   a*-x2   +_^_  =  _V¥Z^2+       «2       , 
y/a2-x2         Va2-x2      y/a2-x2  \fa2-x2 

We  have  therefore 

f  Va2  -x2dx  =  xy/'a2  -x2-  f  Va*  -  x2  dx  +  «2  f — — — . 
•>  J  JVa2-z2 

whence        2  (  Va2  -  x2  dx  =  x  Va2  —  x2  +  a2  arc  sin  - , 
J  a 

and  f  Va2'—  x2  dx  =  -  \xVa2  -  x2  +  a2  arc  sin  - 1  • 


Arts.  114,  115]     INTEGRATION   OF   RATIONAL  FRACTIONS        209 

EXERCISES 

1.     \x*exdx.  2.  fx'n  log  x  dx. 

3.     (  arc  sin  Odd.  4.  j  arc  cot  6 dd. 

5.     \sin2ddd.  6.  (eax  sin  bxdx. 


7.  j  log  x  dx.  8.  (cos  d  log  sin  6  dd. 

9.  f  x2  arc  tan  x  dx.  10 .  f  e a  cos  -  dx. 

11.  fxcossedx.  12.  fsec80d0. 

13.  CVx*  +  a*dx.  14.  §*x»y/a*-x*dx. 

15.  (\b8  log2  a  rife.  16.  f* 


cos  0  cos  2  0<20. 


115.  Integration  of  rational  fractions.  By  a  rational  fraction 
we  mean  the  quotient  of  two  rational  integral  functions.  We 
shall  consider  only  those  fractions  in  which  the  numerator  is  of 
lower  degree  than  the  denominator;  for,  if  this  is  not  the  case, 
we  can  always  by  division  reduce  the  given  expression  to  a  ra- 
tional integral  function  plus  such  a  fraction.  The  decomposition 
of  rational  fractions  is  fully  treated  in  algebra,1*  and  a  knowledge 
of  the  principles  and  methods  involved  is  assumed  here.  In  the 
present  article,  we  shall  show  how  the  decomposition  of  rational 
fractions  may  be  employed  in  simplifying  an  integration.  We 
shall  consider  the  following  cases. 

(a)  When  the  denominator  is  the  product  of  several  linear  factors, 
none  of  which  is  repeated. 

The  given  function  can  then  be  written  in  the  form 

ffx\  = 4M 

We  may  now  assume 

/(x)=^-  +  -B-+  ...  +^-, 
x  —  «!      x  —  a2  x  —  ak 


*  See  Rietz  &  Crathorne's  College  Algebra,  pp.  203-208. 


210  SPECIAL  METHODS   OF  INTEGRATION       [Chap.  XII. 

and  calculate  the  numerators  A,  B,  •••,  K  by  the  principles  of 
undetermined  coefficients.  The  integral  of  f(x)  is  then  found 
by  taking  the  sum  of  the  integrals  of  the  separate  terms.  The 
following  examples  illustrate  the  method. 


Ex   !       f  (5s  +  l)<fa 
J  &  _  2  x  -  86 


We  have 

5x  +  1  5x  +  1  A  B 


x2_2x-35      (x+5)(x-7)      x  +  5      x-7 

Clearing  of  fractions,  we  obtain 

5  x  +  1  =  A  (x  —  7)  +  B  (x  +  5). 

Equating  the  coefficients  of  the  same  powers  of  x  in  the  two  members  of  this 
equation,  there  results 

5  =  A  +  B, 
and  1  =  5  B  -  7  A, 

from  which 

A  =  2,    B=S. 

Substituting  these  values  in  the  original  equation,  we  have 

C    (5  x  +  1)  dx    _  C'2dx        r3dx 
J(x  +  5)(x  —  7)      J  x  +  5      Jx  —  7 

=  2  log  (as  +  5)  +  3  log  (x  -  7)  +  C 

=  log[(x+5)2(x-7)3]  +  C. 


Ex.2. 


J    a3  +  xa  _  6 


—  1)  dx 
x 


Since  ^+  x2  -  6x  =  x(x  4-  8)(«  -  2), 

™  k—  x2  +  x  - 1  A  ,      B  C 

we  have  ! =  — -\ • 

x(x+3)(x-2)       x      x+3     x-2 

Clearing  of  fractions,  we  get  the  identity 

x2  +  x  -  1  =  A(x  +  3)  (x  -  2)  +  Bx{x  -  2)  +  Cx(x  +  3). 

For  x  =  0,  we  obtain  A  =  £ ;  f or  x  =  —  3,  B  =  £  ;  and  for  x  =  2,  G 
Hence,  we  have 

ga  +  s-l    =  1  l  l 


xS  +  xS-Gx      6x      3  (x  +  3)      2  (x-2) 

Therefore,  f  (*2  +  *  -  1)  <*x  =  fjfc  +  f      dx  C      dx 

'J    X3  +  X2_6x        Jcx      J3(x+3)      J  2(x-2) 

=  J  log  x  +  I  log  (x  +  3)  +  i  log  (x  -  2)  +  0 
=  log  \/x(x  +  3)2(x-2)3  +  (7. 


Art.  115]       INTEGRATION   OF   RATIONAL   FRACTIONS  211 

EXERCISES 

±     C  (x  +  2)  dx  |  2     C  (3  as  -  5)  dx 

'  Jx*-bx  +  Q  Jx2  +  2x-  15* 

3     f    (3x  +  4)dx  .     4     C(2x-l)dx 

J  xs  -  7  x2  +  12  x'  J  x'2  +  3  x  +  2 ' 


J2+7x+3x2  J      X3_4a.2_5a. 


r."  f-I* 

J  x(x  + 


+  mn)  dx  8     r  9  x2-  4  x  -  8  - 

m)  (x  —  n)  J       xs  —  4  x 


(6)    Wftefi  £/ie  denominator  is  the  product  of  several  linear  factors, 
some  of  which  are  repeated. 

In  this  case,  the  given  function  has  the  form 


/(*)  = 


*(s) 


(a;—a)(x  —  P)*:>  (x  —  y)'' 

which  can  be  written  in  the  form 


»-y    0»-y)2  0-y)3 

We  may  calculate  the  coefficients  A,  B,  C,  •••,  K,  L,  M, •••,#, 
in  case  (a).     The  following  example  illustrates  this  method. 


Ex.      f     (x-S)dx 
J  x*  —  4  x2  -(-  4  x 


We  have 


_4  .      Jg      . 


X3  _  4  X2  +  4  x      x  (X  _  2)2      a;   '  a;  _  2      (x  -  2)2 
Clearing  of  fractions,  we  get  the  identity 

x  -  8  =  A  (x  -  2)2  +  Bx  (x  -  2)  +  Cx. 
Equating  the  coefficients  of  like  powers  of  x,  we  have 

-8  =  4  A,         1  =  -4A-2B+C,        Q  =  A  +  B 
and  from  these  equations  we  obtain 

A  =-2,         B=2  C=-3. 


212  SPECIAL   METHODS   OF  INTEGRATION       [Chap.  XII. 


Hence  f.     M^^jfil  +  I  f-*L_  -  3  f      ** 

Jx3-4«H4x  Ja;  J  a  -  2        J  (x  -  2)2 

=  _iog^  +  iog(x_2)2+_§_  +  a 


EXERCISES 


1     f     yx  -  4)  c?x  2     f  (3 a;  -  2)  da; 

J  xa  -  4  x2  +  4  x '  J    x(x  +  3)2 

3      f  (2  x  -  5)  dx  4      f  (x-  l)c?x 
J      (x  -  2)3  J    (x  +  3)2  ' 

5      f(4y-3)dy  6      f  6X3  -  8  x2  -  4x  +  1^ 
'*  J     v3-3i/'2  'J 


y8-3y2  J        x4  -  2  x8  +  xs 

x-1) 


_      f    (3  x  —  1 )  dx  q     C  xdx 

J  xa-x2-X+\'  '    J  (x+  1)2( 


fx2  +  2x-3^  r 

J       X3  -  x2  J  ( 


mx2dx 
w  +  x)* 


(c)    W/ien  £/*e  denominator  contains  factors  of  the  second  degree. 

Aside  from  linear  factors,  the  denominator  may  contain  quadratic 
factors  not  decomposable  into  real  linear  factors.  For  the  linear 
factors,  we  assume  a  decomposition  into  partial  fractions  in  ac- 
cordance with  the  principles  of  cases  (a)  and  (b).  Corresponding 
to  the  quadratic  factors,  we  assume  fractions  whose  numerators 
are  linear  in  the  variable.  If  any  of  the  quadratic  factors  appear 
to  a  degree  higher  than  the  first,  then  we  assume  in  the  decompo- 
sition as  many  fractions  as  the  degree  of  the  factor,  the  numera- 
tor of  each  being  linear  and  the  denominator  having  the  given 
factor  in  increasing  powers.  The  following  examples  illustrate 
the  method. 

Ex.1.    <"         rMx 


J  (x 


l)(x2  +  l) 

We  write  * =  -A-  +  Bx  +  C. 

(x-l)(x2  +  l)     x-1       x2  +  l 

Clearing  of  fractions  and  equating  coefficients  of  like  powers  of  x,  we  find 


Arts.  115,  116]     FUNCTIONS   CONTAINING  RADICALS  213 

Hence  f *** =1  f_*L  +1  f  «±1  rf* 

J  (a;  -  l)(x2  +  1)      2  J  x  -  1      2  J  x2  +  1 

=  I  log  (x  -  1)  +  \  log  (a:2  +  1)  +  \  arc  tan  x  +  0 

=  |  log  (x  -  l)2  (x2  +  1)  +  \  arc  tan  x  +  C. 

Ex.  2.    f ^±-f (to. 

J  (x-  l)(x2  +  l)2 


If  we  write 

x  +  1  _■   A         Bx+C      Dx+  E 

(x-l)(x2  +  l)2     x-1      x2  +  l       (z2  +  l)2' 
we  find  by  clearing  of  fractions  and  equating  like  powers  of  x  the  following 
values  for  the  coefficients  : 

A  =  \,  B=-\,  C=  -1,D=  -1,  E  =  0. 
Hence 

dx 


J  (x-l)(x2+l)2  2  J  x-1      2Jx2-}-l  J  (x2  +  l)2 

=  |  log  (x  -  1)  -  |  log  (x2  +  1)  —  £  arc  tan  x  + (-  C. 

2  (a;2  +  1) 

EXERCISES 
1.     f^+J°*Z^<fc,  2.    f-*_. 

J  X4  -  1  J  X4  -  1 

3      f(2x4-  l)dx  4      f        xdx 

J   x^(l+x2)2  '  Jsb*  +  4*2  +  8* 

5    r      xdx  6    r  p2  - 1)  dz  t 

J  x4  +  5  x-'  +  4  J  24  +  2  z2  +  1 ' 

7      f  (y3-*)d>j  r  dx 

Jy*  +  2y*-3  '    J  (a  +  a)  (x*  +  62) 


9.     f— * 10.     f- 

J  x2(x2  +  3)  J  x 

11.     f        (8^-D^ 12.     f_ 

J  (l  +  x)2(l  +  x  +  x2)  Jx(l  +  2x2) 


x2  +  x  +  1 
dx 


is.  r  **»  . 

J  (x2  +  l)2 

116.  Integration  of  functions  containing  radicals.  In  the  pres- 
ent article  we  shall  discuss  some  special  methods  by  which 
functions  containing  radicals  may  frequently  be  changed  into 
equivalent  functions  free  from  radicals. 

(a)  When  f{x)  contains  fractional  powers  of  a  +  bx,  but  no  other 
radicals. 


214  SPECIAL   METHODS   OF   INTEGRATION      [Chap.  XII. 

In  this  case,  f(x)  can  be  transformed  into  a  rational  expression 
by  the  substitution 

a  +  bx  =  zn,  (1) 

where  n  is  the  least  common  multiple  of  the  denominators  of  all 
the  fractional  exponents  of  a  +  bx.  This  follows  from  the  fact 
that  /(a?)  and  dx  can  then  be  expressed  rationally  in  terms  of  z 
and  dz,  as  the  following  examples  will  illustrate. 

Ex.1.    Find    f     y*y     . 

(1  +  80* 
Put  1  +  y  =  z3, 

whence  y  =  z3  —  1 ,   dy=S  zHz. 

Wehavethen  f     Eft      =  fg=  *)8  **»  =  3  f(t*-  ,)  dz 


=  s(z4dz-s(zdz 


i)*-i(y  +  i)*+  c. 


Ex.  2.     Find 


p 


^  +  x*  +  4)  dx 


x2  +  1 

Put  x  =  24,  whence  dx  —  4  23dz. 
We  have  then 


f(s*  +  s*  + 


4)  dx  =  f(z2  +  g  +  4)4g3(fe 
J  s2  +  l 


x-  +  1 

/•  /  ft  •  _  i\ 


=  4  [}-  z4  +  $  *3  +  f  z2  -  2  -  |  log (««  +  1)  +  arc  tan  z  +  O] 
=  z4  +  |  z3  +  ^  z2  -  4  0  -  6  log  (z2  + 1)  +  4  arc  tan  z  +  C" 
=  x  +  f  x4  +  6  x*  -  4  x^  -  6  log  (x^  +  1)  +  4  arc  tan  x^  +  C . 


(b)    Whenf(x)  contains  the  radical  Var*  +  ax  +  b  and  no  other. 

In  many  cases  the  integral  may  be  made  to  depend  upon  one  of 
the  fundamental  integrals  by  writing  the  radical  in  the  form 
Vw2  ±  a2.    If  this  method  does  not  lead  to  a  convenient  way  of  per- 


Art.  116]  FUNCTIONS   CONTAINING   RADICALS  215 

forming  the  integration,  we  may  rationalize  f(x)  dx  by  the  substi- 
tution 

■y/x2  +  ax  +  b  =  z  —  x.  (2) 

Squaring  (2),  we  have 

x2  +  ax  +  b  =  z2  —  2  zx  +  x2, 

z2  —  b 

whence  a;  = — ; 

a  +  2z 

ds=2(f  +  ^  +  *)<fe. 

(a  +  2  z)2 

Consequently,  f(x)  dx  becomes  free  from   radicals  upon    substi- 
tuting these  values  of  Vo2  -\-ax-\-b,  x,  and  dx. 

Ex.     f'  +  ^  +  '  +  'fe. 


>f 


2  VP+IT+T 


2Vx2  +  x  +  l  J  V(2x  +  l)2  +  3 

Putting  2  x  +  1  =  u,  we  have  for  the  first  integral 

f  xdx  _  1  f     wriw      _  1  f      du 

J  V(2z-f  l)2  +  3      4^  Vw^TlJ      4*^  Vw2  +  3 


=  I  Vw2  +  3  -  I  log  [«  +  Vm2  +  3]  +  C. 
Replacing  the  value  of  u  and  combining,  we  obtain  as  the  final  result 

\\x+  Vz2  +  x  +  ij  -  \  log  [2  x  + 1;+  2V«2  +  ic+i]  +  a 


(c)    TP#en  /(#)  contains  the  radical  v  —  x2  +  ax  +  b  and  no  other. 

We  shall  consider  only  those  cases  in  which  the  expression 
under  the  radical  can  be  broken  up  into  real  linear  factors.  The 
integration  can  frequently  be  most  readily  performed  by  putting 
the  radical  in  the  form  Va2  —  u2  and  making  use  of  one  of  the 
standard  forms.     If  this  is  not  feasible,  we  may  proceed  as  follows. 

Write  the  radical  in  the  form 


V—  x2-\-ax  +  b  =  V(x  —  a)(fi  —  x), 
and  f(x)  dx  can  be  rationalized  by  the  substitution 


V(a?  —  a)(fi  —  x)  =  (ft  —  x)  z  [or  (x  —  a)  z], 


216  SPECIAL   METHODS   OF   INTEGRATION      [Chap.  XII. 

as  we  shall  now  show.     Squaring,  we  obtain 
(x  -a)(fi-  x)  =  ((3-x)2z2, 

whence  x  =  &±*,  dx  =  ^~a^zdz. 

1+Z2'  (1  +  z2)2 

The  integrand  which  results  from  the  substitutions  is  rational. 
dx 


Ex.     Find 


JxV^ 


xV—x2  +  4:X  —  3 
We  have  here  V(se  —  1)(3  —  x)  =  (8  —  x)  z, 

3z2  +  l      ^_     tedz 


whence  x  =         "*"    ,     dx 

l  +  g2   '  (1+^)2 

Substituting  these  values  in  the  given  integrand,  we  have 

f g  =2r-i^-  =  J-arctanV3^(7 

JxV-x2  +  4x-3        J30-+1       V3 

=J_  arc  tan  JIIEEI)  +  a 
V3  V  3-aj 

(d)  Integration  by  trigonometric  substitution. 

Iif(x)  contains  a  radical  of  the  form  Va2  ±  u2,  where  u  is  some 
function  of  x,  it  can  often  be  easily  transformed  to  an  equivalent 
function  free  from  radicals  by  the  substitution  of  a  trigonometric 
function.  All  that  is  necessary  is  to  substitute  for  u  that  trigono- 
metric function  which  renders  the  expression  under  the  radical 
a  perfect  square.  It  will  be  readily  seen  that  this  end  is  accom- 
plished by  the  following  substitutions  : 

(1)  Put  u  =  a  sin  6,  or  a  cos  0  in  functions  involving  Va2  —  u2. 


(2)  Put  u  =  a  tan  6,  or  a  cot  6  in  functions  involving  Va2  +  u2. 

(3)  Put  u  =  a  sec  0,  or  a  esc  0  in  functions  involving  -\/u2  —  a2. 
Whenever  the  resulting  trigonometric  integrand  does  not  fall 

at  once  under  one  of  the  fundamental  formulas,  or  take  one  of 
the  special  forms  discussed  in  the  following  article,  the  student 
should  apply  the  methods  of  integration  discussed  in  the  preced- 
ing articles. 

The  following  examples  illustrate  the  use  of  these  substitutions. 

Ex.  1.    J  Va2  -  x*  dx. 

Put  x  =  a  sin  z,  dx  =  a  cos  z  dz. 


Art.  116]  TRIGONOMETRIC   SUBSTITUTION  217 

We  have  then 

f  Va?  —  x2  dx  =  a2  \  cos2  z  dz  =  \  a2  f  (1  +  cos  2  z)  dz 

=  \  a2  f  dz  +  \  a2  f  cos  2  z  d(2  z) 

=  \a2z  +  \a2  sin  2z+  C 

=  \a2  arc  sin  -  +  \  xVa?  -  x2  +  C. 
a 
dx 


Ex.  2. 


J  ax 

x2  (x2  -  a2)* 


Put  x  =  a  sec  z,  dx  =  a  sec  0  tan  z  dz. 

Then  we  have 

dx  r  a  sec  2  tan  z  dz 


f- 

Jx2 


s-J 


(x2  -  a2) 2      ^  a2  sec2  2  (a2  sec2  z  -  a2)^ 


=  lfcos^^s-H^+e=Va;2~a2+a 

a2  J  a2  a2x 

2  Vx  dx 


aJx 
EXERCISES 


!-J; 


x3  *MVx-3x* 

dx 


Jl  + 

3.     fx2Oix  +  &)*dx.  4.     f  dX 

J  J  V4  x  -  3  -  x2 

/»  3dx  r  dx 

J  V7x-10-x2'  '    J  Vx2  +  5  x  -  3 

r      <fa  8   c        dx    , 

J  Vx2  -  7  x  +  4  '    J  x  V-  x2  +  5  x  —  6 

r   xs  dx  r(Hx—  1)  dx 

J  V^~l'  '   J     Vx^3 

dx  .   •  c    ofldx 


5 
7 

9 
11 


15. 


17. 


19. 


r          dx  r    x»m 

J  Vx2  +  2  x  +  5  '    J  vT^ 

f-^^4-  14.     p£»E?fc 

J  (1  +  *2)*  J         x'2 

f— *— .  16.     f       ^ • 

J  v/x2  +  x  J  x2Va2  — x2 

r     z2ax  „„     /»     s2<?x 

i r  is.  j—    —3 

J  (a2-x2)*  J(x2-a2)2 

f    ^fe     .  20.     f(x2-a2)*dx. 


218  SPECIAL   METHODS    OF   INTEGRATION       [Chap.  XII. 

r  x*dx  22     r  &&*■   . 

JvO^'  '    J  Vl+x2 


23 


VI -x2 
dx 


Jx^Vx2-! 

24.  fV2rtx-x2o"x.  [Put  x  =  a  (1  +  sin  0).] 

25 .  f 4 — .  [Put  x  =  sec  e.  ] 

Jlx4\/x2-  1 

26.  Show  that  by  the  substitution  of  x  =  a  tan  0  the  integral  f — 

r  J(a2  +  x2)« 

is  reduced  to  an  integral  of  the  form   i  cosp  6  dd. 

27.  Integrate  Nos.  5,  6,  7,  11,  15  without  rationalization. 
Derive  the  following  integrals. 

28.  (    x2dX     =  -^(8a*-4abx  +  3bW)Va~+bx~. 
J  Va  +  6x     15  &3 

29.  fxVa  +  bx  dx=-  -JL  (2  a  -  3  to)  (a  +  bx)%. 
J  15  o2 

30.  fx2  Va  +  6x  ax  =  — ^—  (8  a2  -  12  a&x  +  15  62x2)  (a  +  6x)i 
^  105  o8 

117.  Integration  of  special  trigonometric  functions.  There  are 
certain  types  of  trigonometric  functions  that  are  readily  inte- 
grated by  easy  reductions  to  standard  forms.  The  following  are 
of  this  kind: 

(a)     f  sec2M  oo  doc,         f  csc2w  oc  dx. 

Here  n  is  assumed  to  be  a  positive  integer.  By  the  substitution 
sec2  x  =  1  +  tan2  x,  sec2n  x  becomes  (1  4- tan2  x)M_1  sec2  x ;  hence,  we 

have  I  sec2n  x  dx  —  I  (1  +  tan2  a?)"-1  d  (tan  a;). 

Similarly,        j  csc2n  a;  da;  =  —  J  (1  +  cot2  x)n~l  d  (cot  a;). 

In   each   case  the   binomial  can  be  expanded,  and  the  separate 
terms  of  the  integrand  are  then  standard  forms. 

Ex.  1.      I  sec4  x  dx  =  f  (1  +  tan2  x)  sec2  x  dx 

=  (  sec2  x  dx  +  \  tan2  a  sec2  x  dx 
=  tan  x  +  J  tan3  x  +  C. 


Art.  117]  TRIGONOMETRIC   FUNCTIONS  219 

Evidently  integrals  of  the  form 

j  tanw  x  sec2"  x  dx,  I  cotm  x  sec2"  x  dx 

can  be  integrated  equally  well  by  this  device. 

(b)  f  secm  x  tan2M+1  oc  d<c,  f  cscm  oc  cot2n+ 1  oc  dx. 

Here  n  is  a  positive  integer  (or  zero),  and  m  is  any  number. 
We  have 

J  secm  x  tan2n+1  x  dx  =  j  see"1-1  x  •  (sec2  x  —  1)"  sec  x  tan  x  dx 
=  J  see"1-1  x  •  (sec2  x  —  1)M  d  (sec  #), 

a  form  that  is  readily  integrated.    A  similar  form  may  be  deduced 
for  the  second  integral. 

Ex.  2.      (see4  xtan3  x  dx  =  (  sec3  x  tan2  x  tan  x  sec  £  dx 

—  \  sec3  a- (sec2  x  —  1)  d(sec  x) 

=  i  sec5  a*  cZ(sec  x)  —  (  sec3  x  d(sec  x) 
=  £  sec6  x  —  £  sec4  x+  G. 

(c)  f  sin2w+1  a?  <fcc,  f  eos2n+1  a?  to. 

Again  n  is  assumed  to  be  a  positive  integer.     We  have  in  this 


case, 


j s in2«+l  jj.  ^  _    j    (1  _  cos2  ^n  sjn  x  fix=  _    |£|_  _  cog2  xy  ^(cOS  #). 

j  cos2n+1  xdx=  I  (1  —  sin2  x)n  cos  xdx=  I  (1  —  sin2  #)n  d(sin  #) . 

Either  of  these  may  now  be  integrated  by  expanding  the  paren- 
thesis and  integrating  the  result  term  by  term. 

Ex.  3.      j  sin5  x  dx  =  I  sin4  x  sin  x  dx 

=  —  j  (1  —  cos2  x)2  d(cos  x) 

=  —  Id  (cos x)  +  2  f  cos2 x d(cos x)—  j  cos4 x <Z(cos x) 
=  —  COS  X  +  |  cos3  x  —  i  cos5  X  +  C. 


220  SPECIAL   METHODS   OF   INTEGRATION      [Chap.  XII. 

(d)  \  sin2n  x  dx,        \  cos2w  x  dx. 

We  may  perform  the  integration  in  this   case  by  the  use  of 
multiple  angles.     From  trigonometry  we  have 

cos2  x  =  -i-  (1  +  cos  2  x), 

sin2  x  =  i  (1  —  cos  2  x). 
By  the  aid  of  these  formulas  the  integrand  can  be  expressed  in 
terms  of   multiple  angles  of  the   variable,  where  the   sine   and 
cosine  appear  only  to  the  first  degree. 

Ex.  4.      i  cos  x*  dx. 

We  may  write  this  as  follows  : 
f  (cos2  x)2  dx  =  J  f  (1  +  cos2x)2dx  =  \  f  (1  +  2  cos  2  x  +  cos22  x)  dx 

=  ;*£  +  $  sin  2£-f-i|  (1-f  cos  4  x)  dx 

=  I  x  +  \  sin  2  x  +  \  x  +  -g\  sin  4  as  +  C 
=  |fl5  +  ^sin2x  +  ^  sin 4  as  +  C. 

(e)  f  sinm  x  cosn  a?  to. 

There  are  three  cases  that  demand  consideration. 

(1)  When  either  m  or  n  is  a  positive  odd  integer. 

In  this  case,  we  can  reduce  the  integral  to  the  fundamental 

form    I  undu.     For  example,  suppose  n  is  odd,  say  equal  to  2  fc+1. 

The  given  integral  may  then  be  written 

I  cosm  x  sin2*  x  sin  x  dx  =  —  I  cosm  x  (1  —  cos2  x)kd  (cos  x), 

which  may  be  easily  integrated  by  expanding  the  parenthesis  and 
integrating  the  result  term  by  term. 

(2)  When  m  +  n  is  a  negative  even  integer. 
In  this  case,  we  may  write 

/smmxcosnxdx  =  f  smm  g  cos™+"  x  dx 
J  cosm# 

=  I  ta,nm  x  sec-(m+n)  x  dx. 

Since  —  (ra  -f  ?i)  is  by  hypothesis  a  positive  even  integer,  the  inte- 
gral falls  under  type  (a). 


Art.  118]  USE   OF  TABLE   OF   INTEGRALS  221 

Ex.  5.     f  ^-^dx  =  (ta,n2xsec2xdx  =  ^tan3a:  +  C. 
J  cos4  a;  J 

(3)      When  m  and  n  are  positive  and  both  even. 

We  may  then  reduce  the  given  integral  to  an  integral  depend- 
ing upon  the  integration  of  the  sine  of  some  multiple  of  the  given 
angle.  This  transformation  may  be  accomplished  by  the  aid  of 
the  substitution  sin  2  x  =  2  sin  x  cos  x 

and  those  given  in  (d). 

EXERCISES 


4 


.     fsec6  x  tan  x  dx.        2.     Jcsc4xdx.  3.     (< 

rcos^dXt  5    rsin3  x  cog2  x  dx         6    C 

J  sin4  x  J  J  i 


cot2  x  esc4  x  dx. 

f_.-_9 o  ..  j„  ~  dx 

sin3  x  cos  x 

7.     f  tan3  x  dx.  8.     (  sec2  x  tan5  as  das.  9.     j  sin  x  cot3  x  dx. 

10.     f  ^S15na;(to:  11.     (&m$xcos*xdx.         12.     f  cot2  ac  sec4  x  dx. 

cos5  a: 

118.  Use  of  a  table  of  integrals.  The  process  of  integration  is 
much  more  involved  and  difficult  than  that  of  differentiation.  It 
is  not  possible  to  integrate  all  functions  in  terms  of  elementary 
functions,  and  frequently  when  integration  is  possible  it  is  not 
easily  effected.  Further  consideration  of  the  more  complicated 
cases  is,  however,  rendered  unnecessary  in  an  elementary  text 
because  of  the  existence  of  tables  of  integrals  arranged  for  con- 
venient reference.  Students  should  become  familiar  with  such  a 
table  as  a  labor-saving  device.*  The  table  should  not  be  em- 
ployed, however,  until  the  student  is  thoroughly  familiar  with  the 
various  elementary  processes  of  integration.  The  following  ex- 
amples will  illustrate  the  use  of  such  tables. 

Ex.  1.  Find  the  area  of  a  surface  of  revolution  given  by  the  definite 

integral  2  wa  f"       xdx   -  • 

j0  V2ax-x* 


*  A  good  table  for  this  purpose  is  B.  O.  Pierce's  Short  Table  of  Integrals, 
published  by  Ginn  &  Co.  The  student  is  advised  to  provide  himself  with 
such  a  table. 


222  SPECIAL   METHODS   OF   INTEGRATION       [Chap.  XII. 

From  a  table  of  integrals,  we  find 

r       xdx . -_V2qg-a;2  +  aarcBing-=-g; 

J  y/2ax  —  x2  a 

hence,  the  required  area  is 

1  =  7ra2(7r  -  2). 


2  ira  |  -  V2  ax  -  x2]°  +  a  arc  sin  - — -T  1  = 


Ex.  2.   The  tractrix  is  a  curve  having  a  constant  length  of  tangent.     Re- 
quired the  equation  of  this  curve. 

Denoting  by  a  the  length  of  the  tangent,  we  have  (Art.  39)  a  =  y  esc  0, 


whence  tan  <f>  =  %L  =  ±        g        ,  and  ±  dx  =  ^gEZJE, 

dx         Va2  -  y2  V 

From  the  table  of  integrals, 

C^a2-y2dy=V^Z—2_al     a  +  Va2-y2; 
J        y  y 

hence,  the  required  equation  is 


±  x  +  c  =  V^^V2-a\oga+Va2  ~  y\ 

y 

Ex.  3.   Evaluate  f_^_. 
J  sin4  6 

From  the  table  of  integrals,  we  find 

C    dx    _  _      1  cos  a;      ,  m  —  2  C      dx 

J  sinm  x         m  —  1     sin"1-1  x      m  —  1 J  Bin"-2 

Hence  f-*-  =  -  1  °2i«  +  ?  f_^_. 

J  sin4  0  3  sin8  0      3  J  sin2  0 

But  f  _^L  -  fese2  dd0  =  -  cot  0. 

J  sin2  6     J 

The  required  integral  is,  therefore, 

_  1  (9™1  +  2  cot  flV-Icot  6>(csc2  6  +  2). 
8\sin»0  /         3  v  y 

EXERCISES 

Evaluate  the  following  integrals  by  reference  to  a  table  of  integrals : 

1.     (V2*  sin  xdx.  2.     f ^ -. 

J  x{x2  -  a2)* 

3.     f  "(a2  —  x2)^  dx.  4.     f  sin4  0  cos2  6  dd. 


Art.  119]  APPROXIMATE    INTEGRATION  223 

5.     C' -***—.  6.     f_^ 

Jo  V2  ax  -  x2  J  3  + 


2  cos  a: 
dx 


C        dx  8      C 

J(l+2x2)3  ^x2(3-2a:) 

f dx ^  10      f  xcfx 

J(4-3x  +  x2)3'  '    J(x2  +  2x~Z>y 

n 

1X^     rshixdx^  12_     C2  e  sin  6  cos  6  dd. 


13. 


f  s^njcdx  22      f  2  q 
J      xs  '    J° 

w  » 

f 2  sin7  0  dd.  14.     f2  - — 

Jo  Jo  l  + 


dx 


2  sin  a; 


15      C         x*dx         m  16      (V-x2+6x-ldx. 


17. 


Vx2  -  3  x  +  7 

fviEifc  is.  |      j 

J        x  ^  V5  -  4  x  +  2  x2 


19.     f — -•  20.     f  sin2  6  cos2  0  <ZA 

J  (5  _  4x  +  2a;2)2  J 

119.    Approximate  determination  of  integrals.     As  stated  in  Art. 

102,  the  definite  integral    1    f(dx)  dx  is  represented  graphically 

by  the  area  between  the  curve  y=f(x),  the  X-axis,  and  the  ordi- 
nates  corresponding  to  x  =  a  and  x  =  b.  Ordinarily,  the  definite 
integral  is  evaluated  by  means  of  the  anti-derivative  of  f{x). 
If,  however,  it  is  impossible  or  inconvenient  to  find  the  anti- 
derivative,  the  definite  integral  may  be  evaluated  by  the  following 
method : 

The  curve  y  =  f(x)  is  plotted  from  x  =  a  to  x  =  b,  and  the  area 
between  the  curve,  the  X-axis,  and  the  end  ordinates  is  determined 
approximately  by  one  of  the  methods  described  in  the  following 
articles.     The  numerical  measure  of  the  area  gives  approximately 

the  value  of   J    f(x)  dx.     Since,  as  has  been  shown,  the  definite 

integral  may  represent  any  measurable  magnitude,  as  area, 
volume,  length,  force,  quantity  of  heat,  etc.,  these  methods  may 
obviously  be  used  for  the  approximate  determination  of  any  such 
magnitude. 


224 


SPECIAL   METHODS   OF   INTEGRATION      [Chap.  XII. 


120.    Simpson's  rules. 
M 


^X 


The  problem  of  finding  any  plane  area 
reduces  to  the  problem  of  finding 
the  area  between  a  plane  curve, 
an  assumed  X-axis,  and  two  end 
ordinates.  Thus  in  Fig.  68,  the 
area  of  the  closed  figure  AMBNA 
is  found  by  subtracting  the  area 
between  the  curve  ANB  and  OX 
from  the  area  between  AMB  and 

Fig.  68.  OX. 

Let  ACE,  Fig.  69,  be  a  part  of  a  curve,  and  suppose  the  area 
between  it  and  the  line  XR  is  required,  the  end  ordinates  being 
AX  and  ER.  Let  the  dis- 
tance XR  be  divided  into  a 
number  of  equal  parts,  each 
equal  to  h,  and  through  the 
points  of  division  let  ordi- 
nates 2/j,  y2,  y3,  etc.,  be  drawn. 
Consider  now  the  three  points 
of  the  curve,  A,  B,  and  C. 
Taking  OB  as  the  Y-axis  and 
O  as  the  origin,  the  coordi- 
nates of  these  points  are 

A  =  (-h,  yx),  B  =  (0,  y2),  C  =  (h,  y8).  (1) 

We  now  assume  that  the  equation  of  the  part  of  the  curve  ABC  is 

y  =  a0  +  a1x  +  a2x2.  (2) 

This  is  equivalent  to  the  substitution  of  an  arc  of  a  parabola  with 
a  vertical  axis  for  the  actual  curve.  With  this  assumption,  the 
area  XACP  is 


y  dx  =  |    (a0  +  axx  +  a&?)  dx  =  -  (6  a0  +  2  aji2). 


(3) 


To  determine  the  coefficients  a0  and  a2,  we  substitute  the  coordinates 
given  by  (1)  in  (2).     We  thus  obtain 

V\  =  «o  —  aJl  +  a2^2, 

2/2  =a0,  (4) 

2/3  =  «o  +  a>ih  +  «2^2- 


Art.  120]  SIMPSON'S   RULES  225 

Solving  for  a0  and  a2,  we  get 

ao  —  v*    a2  — 2~p 

and  substituting  these  values  in  (3),  we  have  finally 

area  NABCP  =  |  fa  +  4  y,  +  y8).  (5) 

If  nowwe  assume  the  part  of  the  curve  CDE  replaced  by  a  second 
parabolic  arc,  we  get  in  the  same  way, 

aiea  WM  =  |  (y,  +  4  y4  +  %)i 


and  thus 


area  NAER  =  |  fa  +  4  ft  +  2  &  +  4  y4  +  y5). 


This  process  may  be  repeated  any  number  of  times  ;  hence,  taking 
an  odd  number  of  ordinates  n  -f  1,  dividing  the  figure  into  n  strips 
of  equal  width  h,  the  approximate  area  is  given  by  the  formula 

^=|0/i  +  4i/2  +  22/3+  •••  +2yn_1  +  4yn  +  yn+1).      (6) 

This  result  gives  the  following  rule,  known  as  Simpson's  one-third 
rule: 

Take  the  sum  of  the  end  ordinates,  twice  the  sum  of  the  intervening 
odd  ordinates,  and  four  times  the  sum  of  the  even  ordinates.  Multi- 
ply the  aggregate  by  one  third  of  the  common  distance  between  the 
ordinates. 

If  we  take  four  ordinates  ylf  y2,  y3,  yA,  including  three  spaces,  we 
may  pass  through  their  ends  A,  B,  C,  and  D,  a  third  degree  parabola 

y  =  a0  +  axx  +  a<p?  -f-  asx?. 

Proceeding  as  before  to  determine  the  coefficients,  we  obtain 

area  NADQ  =  ^  fa  +  3 .%  +  3  yz  +  yA).  (7) 

If  therefore  we  divide  the  whole  base  into  some  number  of  parts  n 
divisible  by  3,  and  take  the  space  in  groups  of  three,  or  the  ordi- 


226  SPECIAL   METHODS   OF   INTEGRATION      [Chap.  XII. 

nates  in  groups  of  four,  we  shall  have  for  the  successive  partial 
areas, 

A  =  -£■  (  2/7  +  3  2/s  +  3  y9  +  yM), 


Adding,  we  get  for  the  total  area, 

^=^(2/i  +  32/2  +  3i/3  +  22/44-32/5+  •••  +  Syn  +  yn+1).      (8) 

This  formula  expresses  Simpson's  three-eighths  rule. 

To  show  how  Simpson's  rules  may  be  applied  to  various  magni- 
tudes, we  will  take  the  specific  case  of  the  approximate  determina- 
tion of  the  volume  of  a  solid.  Let  some  line  of  the  solid  be  taken 
as  the  X-axis,  and  let  the  solid  be  cut  by  equidistant  planes  per- 
pendicular to  this  axis.  Then,  if  yi}  y2,  •••,  yn  denote  respectively 
the  areas  of  the  plane  sections,  formula  (6)  or  formula  (8)  gives 

approximately  the  value  of  the  integral    J  y  dx,   taken   between 

proper  limits,  and  therefore  the  volume  of  the  solid. 

Ex.  1.  Find  approximately  the  area  under  the  equilateral  hyperbola 
xy  =  84,  from  x  =  2  to  x  =  8. 

Taking  unit  intervals,  we  have  for  sc  =  2,  3,  •••,  8,  y  =  42,  28,  21,  16.8, 
14,  12,  10.5.     By  Simpson's  one-third  rule, 

A  =  -J  [42  +  10.5  +  2  (21  +  14)  +  4  (28  +  16.8  +  12)]  =  116.67. 

By  the  three-eighths  rule, 

A  =  I  [42  +  10.5  +  2  x  16.8  +  3  (28  +  21  +  14  +  12)]  =  116.67. 

The  exact  area  is  f   y  dx  =  84  f 8  —  =  84  log  -  =  116.45. 

h  *  h   x  &2 

Ex.  2.   Find  approximately  the  value  of  tr  from  the  formula 

^arctanl^1-^-. 
4  Jo  l  +  x2 


Art.  121]  MECHANICAL   INTEGRATION  227 

Here  y  — ,  and  dividing  the  interval  (0,  1)  into  10  parts,  whence 

1  +  se2 

h  =  0.1,  we  get  for  the  successive  ordinates, 

yi  =  l  y5  =  .8620690  y9  =.6097561 

2/2=    .9900990  ?/6  =  .8000000  yl0  =  .5524862 

y3  =    .9615385  y1  =  .7352941  yn  =  .5000000 

y4  =    .9174312  t/8  =  .6711409 

By  Simpson's  one-third  rule,  we  get 

-  =  .78539815,  whence  tt  =  3.141593. 
4 

This  result  is  correct  to  seven  figures. 

Ex.  3.   Find  by  Simpson's  rule  the  volume  of  a  sphere  of  radius  a. 

Dividing  the  diameter  into  four  intervals,  we   have   h  —  ®,  and  y  =  0, 

2 

y  =  |  Tra2,  y  =  ira2,  y  =  ira2,  y  =  £ ira2,  y  =  0.    Hence  by  Simpson's  first  rule, 

4 


Volume  =  I  .  ?  To  +  4  (|  7m2  +  |*a*}  +  2  7ra2l  =  | 


irac 
3 


EXERCISES 

/»19 

1.  Find  \     x2  dx  by  each  of  Simpson's  rules.     Take  h  =  1. 

2.  Find  log  2  approximately  from  the  formula  log  2=  1    ••      Take 

10  intervals  and  use  the  one-third  rule. 

S 

3.  Find  f  3  sin  0  dd.     Take  ft  =  10°  =  — . 

Jo  18 

4.  Air  at  a  pressure  of  40  pounds  per  square  inch  expands  from  a  volume 
of  6  cubic  feet  to  a  volume  of  20  cubic  feet.  The  expansion  follows  the  law 
pv  =  C.  Find  approximately  by  Simpson's  rule  the  work,  done  during  the 
expansion.     Remember  that  the  units  must  be  consistent. 

5.  Find  by  Simpson's  rule  the  volume  of  the  frustum  of  a  cone  whose  base 
radii  are  r\  and  r2  and  whose  altitude  is  h. 

6.  Show  that  the  area  under  the  curve  y  =  kx2  from  x  =  0  to  x  =  a  repre- 
sents the  volume  of  a  cone  of  altitude  a. 

121.  Mechanical  integration.  Instruments  called  mechanical 
integrators  or  planimeters  have  been  devised,  by  means  of  which 
plane  areas  can  be  measured  very  accurately.  A  tracing  point  is 
made  to  follow  the  perimeter  of  the  figure  to  be  measured,  and  the 
area  is  given  by  the  reading  of  a  recording  wheel.     For  a  descrip- 


228  SPECIAL  METHODS   OF   INTEGRATION      [Chap.  XII. 

tion  of  the  planimeter  and  a  discussion  of  its  theory  the  student 
is  referred  to  the  standard  works  on  engineering.*1 

MISCELLANEOUS    EXERCISES 


Integrate  the  following. 

dx  of  dx  o     C  y/%  —  Vx 


dx. 


f        dx  2     C  dx 3     (vx  — 

"*  x2(l  +  a;2)l  J  y/4  +  8z-2x*  J      4X* 

Jdx  5     C     xs  dx  6     f    cosddd 
Va+  Vx                    'JV^^TT*  '*  Jcos(0  +  e)' 

7.  Integrate  in  three  different  ways  each  of  the  following. 

(a)    f  (a2-x2)^dx;  (6)    fx2  Va2  -  x2  dx. 

8.  Show  that  f  tan*  xd  x  =  tanW-1  -  -  f  tan""8  x  dx. 

J  n  —  1       J 

Suggestion.     Use  the  substitution  tan'2  x  =  sec2  x  —  1. 

9.  Making  use  of  the  reduction  formula  of  Ex.  8,  find 

(a)    ftan5xdx;  (6)    ftan40d0. 

Perform  the  following  integrations. 

C     °jX  —  2  C 

10 .     I  — dx.  11.     \  x'2  arc  cos  x  dx. 

Jx2-5x  +  4  J 

/» /j  arc  tan  a;  ^  /• 

12.     y  ^—7'  13      |e2xcos2xdx. 

(1  +  x2)*  J 

14.  Integrate  Cx  ^^  ~  x%  dx  by  the  substitution  x2  =  a2  cos  2  0. 

J   Va2  +  x2 

15.  By  the  method  of  integration  by  parts  deduce  the  following  formulas. 

(a)    f  sin*  x  dx  =  -  sinW"1  x  cosx  +  ^i  f  sin*-2  x  dx  ; 
J  n  «    J 

(6)    f  cos*  x  dx  =  sin  x  C0SW"1  *  +  tLzl  f cos*-2 X  dx. 
J  n  n    J 

16.  Using  the  formulas  of  Ex.  15  find  the  following  integrals. 

(a)    fsin50d0;  (6)    fcos40d0. 


*  See,  for  example,  Johnson's  Suv  eying  or  Carpenter's  Experimental 
Engineering ;  also  the  descriptive  pamphlets  issued  by  firms  manufacturing 
such  instruments. 


Art.  121 J  MISCELLANEOUS   EXERCISES  229 


17.    Prove  that   i     sinwxdx=\     cosnxdx 
Jo  Jo 

1.8-6  ..  (n-  1)     w  «     • 

= — — * J-  .  - ,  it  »  is  an  even  integer, 

2-4-6  •••  n  2'  5    ' 

2.4  -6  ...  O-  1)     .,      .  , ,  .   . 

=  — * L ,  if  n  is  an  odd  integer. 

3.5.7  •••  n 

Evaluate  the  following  definite  integrals. 

IT 


18. 


f-  0sin0cos0d6».      19.    ("  d  sin*  d  cos  Odd.      20     f  2sin4g  $q 
J-|  Jo  '   Jo    sintf 


Integrate  the  following. 

21.    f dx  _•       22.    f— ^L 23.    (Wa  +  &x2(2x. 

•J  (l-x2)Vl+x2  J  (x2+a2)(x2  +  &2)  J 

24.    By  means  of  the  substitution  z  =  tan  -  verify  the  following  results  : 

f * =  — l=arctanfJiEltanl?,  (a2>62), 

Ja  +  &cosx      Va2-&2  L*a  +  6        J 2'  v  " 

VHa  +  V6  -  a  tan  - 
los *i  («2<&2)- 


V&2  -  a2        V 6  +  a  -  V&^tan  5 


dx  2 

arc  tan 


r       dx 
J  a  +  b  sin  x 


Va2  -  62 


a  tan  -  +  b 


Va2 


,  (a2>62), 


-               atan^+6-V62-a2 
=        1         log ? ,  («2<&2). 

25.  Find  the  area  of  the  ellipse  —  +  —  =  1. 

a2      b2 

26.  Find  the  area  bounded  by  the  curve  (-)    +  (-]    =  1« 

27.  Find  the  area  between  the  cissoid  y2  =      x*       and  its  asymptote 
0  ■       2  a  —  x 

x  =  2a. 

28.  Find  the  area  under  the  curve  y  =  \ogx  from  x  =  1  to  x  =  10. 

29.  Find  the  area  of  the  loop  of  the  curve  my2  =  (x  —  a)  (x  —  b)2. 

30.  Find  the  length  (a)  of  the  curve  y  =  ex  from  x  =  0  to  x  —  2  ;  (&)  of 
the  curve  ?/  =  logx  from  x  =  1  to  x  =  6  ;  (c)  of  the  curve  9y2  =  x(x  —  3)2 
from  x  =  0  to  x  =  3. 

31.  Find  the  length  of  the  spiral  p  =  ad  from  the  origin  to  the  point  (ira,  7r). 


230  SPECIAL   METHODS   OF  INTEGRATION      [Chap.  XII. 

sin2  6 

32.  Find  the  length  of  an  arc  of  the  cissoid  p  =  2  a- from  6  =  0  to 

COS0 

a 

33.  Find  the  length  of  the  curve  p  =  asm3-- 

o 

34.  A  circle  of  radius  -  rolls  on  a  circle  of  radius  a,  and  a  point  on  the 

circumference  of  the  rolling  circle  traces  an  epicycloid  whose  polar  equation 
is  4(p2  _  a2)3  —  27  a4/?2  sin2  6.     Find  the  whole  length  of  the  curve  thus  traced. 

A2     v'2 

35.  Find  the  volume  generated  by  revolving  the  ellipse  —•+-*-  =  1  about 

the  X-axis.  .         a2     b* 

36.  Find  the  volume  generated  by  revolving  one  arch  of  the  cycloid 

x  =  a{6  —  sin  0),     y  =  a{\  —  cos  0) 
about  the  axis  OX. 

37.  Find  the  volume  generated  if  the  cycloid  is  revolved  about  OY. 

38.  Find  the  volume  generated  by  revolving  one  arch  of  the  curve  y  =  cos  x 
about  the  axis  OX. 

39.  Find  the  volume  generated  by  revolving  the  cardioid  p  =  2  a(  1  —  cos  6) 
about  the  axis  OX. 

40.  Two  cylinders,  with  the  same  altitude  h,  have  a  common  upper  base 
of  radius  a,  and  the  lower  bases  are  tangent  to  each  other.  Find  the  volume 
common  to  the  two  cylinders. 

41.  Find  the  surface  generated  by  revolving  the  ellipse  —  4-  V-  —  \  about 
its  major  axis  ;  also  about  its  minor  axis. 

42.  Find  the  surface  generated  by  revolving  the  cardioid  p  =  2  a{\  —  cos  6) 
about  the  initial  line. 

43.  Find  the  mean  length  of  the  ordinates  of  a  semicircle  of  radius  «,  if 
the  ordinates  are  taken  at  equidistant  intervals  on  the  diameter. 

44.  Find  the  mean  distance  of  the  points  on  the  circumference  of  a  circle 
of  radius  a  from  a  fixed  point  on  the  circumference. 

45.  Zeuner's  equation  for  superheated  steam  is pv  =  BT  +  Cpn.  For  an 
isothermal  expansion  the  temperature  T  is  constant.  Derive  an  expression 
for  the  work  done  during  an  isothermal  expansion  from  v\  to  v2. 

Suggestion.     Use  the  equation 

work  =  I  p  dv  =  pv  —  \  v  dp. 

46.  The  acceleration  of  a  particle  that  moves  under  the  influence  of  an 

attractive  force  that  varies  inversely  as  the  square  of  the  distance  is  a  =  — - 

s2 

Derive  a  relation  between  v  and  s,  also  between  t  and  s,  taking  s0  as  the 
initial  distance  of  the  particle  from  the  center  of  attraction. 


CHAPTER   XIII 
FUNCTIONS  OF  TWO  OR  MORE  VARIABLES 

122.  Definition  of  a  function  of  several  variables.  Heretofore 
we  have  discussed  functions  of  a  single  independent  variable. 
A  function,  however,  may  depend  upon  two  or  more  variables 
having  no  mutual  relation,  that  is,  independent  of  one  another. 
Thus  the  volume  of  a  gas  depends  upon  the  temperature  and  also 
upon  the  pressure  to  which  it  is  subjected,  and  the  pressure  and 
temperature  may  vary  independently.     In  general,  we  may  say : 

z  is  a  function  of  the  independent  variables  x,  y,  •••  ichen  for  each 
set  of  values  of  these  variables  there  is  determined  a  definite  value  or 
values  of  z. 

A  function  of  two  variables 

»=/(*,  y) 

is  represented  geometrically  by  a  surface,  and  to  each  pair  of 
values  of  (x,  y)  there  corresponds  a  point  on  this  surface.  The 
motion  of  a  point  on  this  surface  depends  upon  the  manner  in 
which  x  and  y  vary.  One  of  these  variables  may  remain  constant 
while  the  other  is  allowed  to  vary,  or  the  two  variables  may 
change  simultaneously.  In  the  first  case,  the  extremity  of  the 
ordinate  describes  a  plane  curve  lying  in  a  plane  parallel  to  the 
YZ-  or  XZ:plane ;  and  in  the  second  case  the  extremity  of  the 
ordinate  may  describe  a  space  curve. 

When  one  of  the  variables  remains  constant,  say  y  =  yn,  we  say 
that  the  function  f(x,  y)  is  continuous  in  x  at  the  point  (x0,  y0)  if 

£    /0>  2/o)=/<>o,  2A>).  CO 

x  =  x0 

Likewise  if  we  have 

l  f(xo,y)=f(%o,  2/0),  (2) 

y  =  yo 

we  say  that  f(x,  y)  is  continuous  in  y  at  the  point  (x0,  y0).     In  order 

231 


232  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES    [Chap.  XIII. 

that  the  function  be  continuous  in  both  variables  together  at  the 
point  in  question  we  must  have 

L    f(x,y)=f(x0,y0).  (3) 

x  =  x0 
y  =  2/0 

As  will  be  seen  from  these  limits,  the  continuity  in  (x,  y)  together 
involves  of  necessity  continuity  in  x  and  in  y.  In  general,  when- 
ever we  speak  of  the  continuity  of  a  function  of  two  variables,  the 
continuity  with  respect  to  both  variables  taken  together  is  to  be 
understood  unless  otherwise  stated. 

123.  Partial  derivatives.  In  the  preceding  article  it  was  pointed 
out  that  a  function  of  two  variables  may  vary  either  by  permitting 
one  of  the  variables  to  remain  constant  while  the  other  changes, 
or  by  allowing  both  to  vary  simultaneously.  The  increment  of 
the  function /(#,  y)  due  to  a  change  in  x  alone  is 

/(aj  +  Aa?,  y)-f(x,y). 

Let  us  consider  the  ratio  of  this  increment  to  the  increment  of 
the  variable  x.     The  limit  of  this  ratio  as  Ax  =  0,  viz. : 

L    /(a?  +  Aa?,  y)-f(x,y) 
Ax  ±  o  &x 

is  called  the  partial  derivative  of  f(x,  y)  with  respect  to  x.  Simi- 
larly,  the  limit  ^     f(x,y  +  Ly)-f{x,y) 

Ay  ~  0  ty 

is  called  the  partial  derivative  off(x,  y)  with  respect  to  y.  These 
derivatives  are  called  partial  derivatives  because  they  measure 
only  partially  the  variation  of  the  function  as  compared  with  that 
of  the  variables.  To  distinguish  these  from  the  derivatives  which 
have  been  thus  far  considered,  we  shall  call  the  latter  total  deriva- 
tives. In  a  subsequent  article  we  shall  discuss  methods  of  deter- 
mining total  derivatives  of  functions  of  several  variables. 

To  distinguish  symbolically  the  two  classes  of  derivatives,  we 
denote  the  partial  derivatives  by  using  the  round  d  instead  of  d. 
Thus  the  partial  derivatives  of  z=f(x,  y)  with  respect  to  x  and  y 
are  written  dz     dz 

dx'   dif 


Art.  123]  PARTIAL   DERIVATIVES  233 

respectively.  They  are  frequently  represented  also  by  the  equiva- 
lent symbols 

/„'(*,  *),'//&  2f> 

Partial  derivatives  usually  involve  both  x  and  y,  and  "may  like- 
wise have  partial  derivatives  with  respect  to  either  variable. 
Thus  we  may  have 

JL(§*\    <L(§*\  JL{<!*\  A.{te\ 

dx\dxf  dy\dyf  dx\dy/    dy\dxj 
These  higher  partial  derivatives  are  represented  symbolically  by 

dh     dh     J2^       d2z 
dx2'    dy2'   dx  by     By  dx' 
or  by 

/."(*,  y),  /,"(*,  y),  /«,"(*,  y),  /„"(*,  y), 

respectively. 

We  may  extend  these  considerations  to  partial  derivatives  of 

still  higher  order  and  to  functions  of  more  than  two  variables. 

The  notation  employed  for  the  partial  derivatives  of  higher  order 

indicates  the  number  of  differentiations  and  the  order  in  which 

d3u 
they  are  made.     Thus,  - — — -  indicates  three  differentiations,  the 
dxdy2 

d5u 
first  two  with  respect  to  y,  the  third  with  respect  to  x :   — ^  „  »  0 

dz  dx2  dy2 

indicates  five  differentiations,  the  first  and  second  with  respect 

to  y,  the  third  and  fourth  with  respect  to  x,  and  the  fifth  with 

respect  to  z. 

Since   the  function   z  =f(x,  y)   is   represented   by   a   surface, 

z  =f(x,  b)  is  the  equation  of  a  plane  curve  cut  from  this  surface 

dz 
by  the  plane   y  =  b.     For  the   derivative  —  we  have  the   same 

dx  dz 

geometric  interpretation  in  this  plane  as  was  given  earlier  for  — > 
namely,  the  slope  of  the  tangent  to  the  curve  in  question. 

Let  z  =f(x,  y)  be  the  surface  shown  in  Fig.  70.  Consider  any 
point  on  this  surface,  as  P,  at  the  intersection  of  the  curves  BPC 
and  EPFj  cut  from  the  surface  by  the  planes  y  =  b  and  x  =  a,  re- 
spectively.     Then  the  slope  of  the  curve  BPC  is  given  by  the 


234  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES    [Chap.  XIII. 


dz 
partial  derivative  —-,  and  that  of  the  curve  EPF  by  the  partial 

dz 

derivative  —     That  is, 

dy 

tan  <£  =  --,   talli- 
ca? dy 

The  values  of  tan  <£  and 
tan  if/  for  some  definite  point 
P  on  the  surface  are  ob- 
tained  by   substituting   in 

dz 
the  expressions  for  —  and 

dz  .  dx 

—  ,  respectively,  the  corre- 
sponding values  of  x  and  y. 
Thus  if  (a,  b)  is  the  projec- 
tion of  P  on  the  X  F-plane, 
we  substitute  a  for  x  and 
b  for  y. 
The  process  of  finding  a  partial  derivative  is  in  all  respects  the 
same  as  that   employed  in  finding  an  ordinary  derivative  of  a 
function  of  a  single  variable.     The  following  examples  illustrate 
this  statement. 

Ex.  1.     Find  —   and  -£,  when  z  =  y2  sin  x. 

dx  dy 

Treating  y  as  a  constant  and  differentiating  with  respect  to  x,  we  have 

dz       Q 

—  =  y1  cos  x. 

dx 
Likewise,  if  we  consider  x  constant  and  differentiate  with  respect  to  y,  we  get 


We  have 


Fig.  70. 


dz 
dy 

=  2  y  sin  x. 

juccessi1 

/e  partial  derivatives  of  the  functi 

z  = 

=  xex  log  y. 

dx 

--(l  +  x)  e* 

logy;  f-*  = 
dy 

xex  _ 

y 

frz_ 
dx2 

(2  +  x)  e*  1 

d^z 

_xex 
y*' 

d*z   _ 

Jl  +  x)  e* 

faz 

(1  +  x)  ex 

dydx 

y 

dx  dy 

y 

Art.  123]  PARTIAL  DERIVATIVES  235 

Ex.  3.     The  surface  z  —  —  +  2-  is  cut  by  planes  x  =  4,  y  =  3.     Find  the 
8      12 

slopes  of  the  curves  cut  from  the  surface  by  these  planes  at  the  point  of 

intersection  of  the  curves. 

We  have  here 

dz  _  x         dz  _y 

dx~^        dy~Q 

At  the  specified  point,  therefore, 


tan 


0  =  ?~|       =  1 ;  tan  xL  =  21      =  I. 


EXERCISES 

Find  the  partial  derivatives  —  and  —  for  the  following  functions. 
dx  dy 

1.    z  =  x2y5.  2.    z  =  sin  x  cosy. 

3.    z  =  ye2  -f  are*.  4.   z  =  xS  —  az2?/  +  &£y2  -f  2/*. 

5.   z  =  yx.  6.    z  =  arc  secu- 

re 

7.   z  =  y  sin  £  +  x  sin  y.  8.   z  =  sin  («  +  y). 

fill  (17/  (J7t 

Find  —  ,  —  ,  — ■  for  the  following  functions  of  three  variables. 
dx     dy      dz 

9.   u  =  3  x?yz~*.  10.   u  =  e*  log  ?/2. 

11.   w  =  sin  a  cos  y  +  sin  y  cos  2  +  sin  z  cos  a\ 

12.  «=iog«±*.'  13.  M  =  *L+^±i!. 

14.  The  volume  of  a  cone  may  be  expressed  as  a  function  of  the  altitude 
and  of  the  radius  of  the  base.  Denoting  the  volume  by  J7,  the  base  radius  by 
r,  and  the  altitude  by  h,  express  V  in  terms  of  r  and  h,  find  the  partial  de- 
rivatives —  ,  — ,  and  give  interpretations  of  these  derivatives. 

dr       dh 

15.  Express  the  area  A  of  a  triangle  as  a  function  of  its  base  x  and  alti- 
tude y.  Find  the  rate  at  which  the  area  changes  :  («)  when  the  altitude  re- 
mains unchanged  and  the  base  varies  ;  (6)  when  the  base  is  constant  and 
the  altitude  varies. 

16.  Express  the  area  A  of  a  triangle  in  terms  of  two  sides  m  and  n  and 
the  included  angle  6.  Find  the  rate  of  change  of  A :  (a)  when  9  changes, 
m  and  n  remaining  the  same  ;  (6)  when  m  changes,  0  and  n  remaining  the 
same. 


236  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES     [Chap.  XIII. 

17.  Interpret  geometrically  the  equations 

dx  dy 

Form  ^,   i^-  ,  i^_,  and  ^  for  the  following  functions. 
dx2     dxdy     dy  dx  dy'2 

18.  u  =  x  log  y.  19.    u  =  xmyn.  20.    u  =  x3  +  axy  —  ys. 
21.   u  =  ex+y.                 22.   u  =  x(l  —  ys). 

23.   The  formula  H=kssD*   is  used  to  determine  the  horse  power  re- 
quired to  drive  a  steamship,  where  s  denotes  the  speed  and  D  the  displacement. 

What  is  the  interpretation  of  the  derivative  - —  ?  of  -r-  ?    Derive  expres- 
sions for  these  derivatives.  **8  oD 


124.    Interchange  of  order  of  differentiation.     Among  the  partial 
derivatives  enumerated  in  Art.  123  were  the  second  derivatives 

-   \  f  ,  \  «     In  most  cases  that  arise  in  the  applications  of  the 
dx  dy    dy  dx 

calculus  to   physical  problems,  it  is  a  matter  of  indifference  in 

what   order  the   differentiation   is   performed ;  that  is,  in  most 

cases  these  two  partial  derivatives  give  the  same  result. 

Let  us  consider  the  limits  involved.     We  have,  by  definition, 

§£=     L    fjx  +  Ax,y)-fjx,  y)  ~v 

dx      Ax  =  0  A« 

V=  l  f(x>y+*y)-f(x>y).  (2) 

&J      Ay  =  0  Ay 

Consequently,  we  have 

f(x + Ax,  y  -f-  Ay)  -fjx,  y + Ay)    f(x  +  Ax,  y)  -fjx,  y) 
&f  _   T         T  Ax  Ax 


dydx  a?/=OAx  =  0  Ay 

_  L       L  Ax  +  Ax,  y  +  Ay)  -f{x,  y  +  Ay)  -/(x  -f  Ax,  y)  +/(x,  y) 
A</  =  OAx=0  Ay  Ax  ,^ 

and  similarly, 

ay  ^  jr       ^  /(x+Ax,y  +  Ay)-/(x,y+Ay)-/(x+Ax,y)+/(x,y) 
dxdy  Ax ±0 Ay ±0  Ay  Ax  ^ 


Arts.  124,  125]  TOTAL   DERIVATIVES  237 

It  appears  that  the  only  difference  between  the  two  derivatives 
is  the  order  in  which  the  increments  Ax  and  Ay  are  allowed  to 
approach  zero.  It  can  be  shown  that  this  order  is  always  a  matter 
of  indifference  if  fj,  fvx"  (or  fy'yfj')  are  continuous  functions  of 
the  two  variables  (x,  y)  taken  together.* 


EXERCISES 

B2u        d2u 


In  each  of  the  following  functions  show  that 

Bx  By     dy  dx 

1.    u  =  xsy2  —  4  xy*.  2.    u  =  cos  (a;  +  y).  3.   u  =  exsiny. 

4.    u  =  yx.  5.   u  =  log  (x2  +  y'2).  6.   u  =  cos  xy2. 

Bzu          dsu 
For  each  of  the  following  functions  show  that  = 

dx  dy2     dy2  dx 
7.    u  =  x2(x  —  y) .  8.    u  =  xy  cos  (x  +  y). 

9.    u  —  y  log(l  +  xy).  10.    u  =  sin2  x  cos  y- 

125.  Total  derivatives.  Let  z  =f(x,  y)  be  a  continuous  func- 
tion of  the  two  independent  variables,  having  continuous  deriva- 
tives ;  and  let  both  x  and  y  be  arbitrarily  chosen  continuous 
functions  of  a  common  variable  t,  having  the  continuous  deriva- 

ii  /v*     it')/  cly 

tives  — ,     -•     We  shall  attempt  to  find  an  expression  for  —  in 
dt'  dt  dt 

(1-1*  flit 

terms  of  —  and  — ,  in  other  words,  to  express  the  rate  of  change 
dt  dt  '  r 

of  z  in  terms  of  the  rates  of  change  of  x  and  y. 

For  the  increment  of  z,  we  have 

Az  =f(x  +  Ax,  y  +  Ay)-f(x,  y), 

which  may  be  written  in  the  form 

Az  =  [f(x  +  Ax,  y+Ay)-f(x,  y-j- A*/)] +  [/(*,  y+Ay)-f(x,  y)\  (1) 

By  an  application  of  the  law  of  the  mean,  we  have 

f{x  +  Ax,  y  +  Ay)  -f(x,  y  +  Ay)  =fx\x  +  Qx  Ax,  y  +  Ay)  Ax,    (2) 
f(x,  y  +  Ay) -f(x,  y)  =fy'(x,  y  +  02  Ay)  Ay,  (3) 


*  See  First  Course,  pp.  247  et  seq. 


238  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES     [Chap.  XIII. 

where  Bx  and  02  lie  between  0  and  1.  Substituting  these  values 
in  the  second  member  of  (1),  we  have,  after  dividing  by  At, 

ft  =/.'(*  +  6,  Ax,  y  +  ^y)~  +/,'(*,  y  +  9,  Ay)  ft  •       (4): 

In  the  limit,  we  have 

L    Af  =  ^j      L    Ax  =  ^      L    Ay  =  dym  (5) 

At^O&t       dt      A£~0A£       dt      fa±Q&t       dt 

Moreover,  since  Ax  and  Ay  approach  zero  with  At,  and  fj(x,  y) 
and  fy'(x,  y)  are  continuous  functions  of  both  variables  together, 

we  have  L    fJ(x  +  0i  Aa;,  y  + Ay) =/,'(*,  y),  (6) 

A£  =  0 

L    fy\x,y  +  e2Ay)=fy\x,y).  (7) 

At  =  0 

Writing  j-,  -i  iovfJ(x,  ?/),  /y'(a?,  2/),  respectively,  we  have  from  (4) 

dz  __  dz  dao      dz  d]/  #  .  . 

eW      da?  eW      62/  <« "  W 

This  result  may  be  expressed  in  words  as  follows : 
TJie  total  rate  of  change  of  a  function  of  x  and  y  is  made  up  of 
two  parts :  one  is  the  rate  of  change  of  x  multiplied  by  the  partial 
x-derivative  of  the  function,  the  other  the  rate  of  change  of  y  multi- 
plied by  the  partial  y-derivative. 

The  following  examples  will  serve  to  illustrate  this  principle. 

Ex.  1.  Suppose  the  characteristic  equation  of  a  gas  to  be  pv  =  54  T,  and 
let  the  volume  and  temperature  at  a  given  instant  be  Vo  =  15  and  T0  =  640. 
The  corresponding  pressure  is 

54_x640  =  2304 
ro  16 

In  this  state,  suppose  the  temperature  to  be  rising  at  the  rate  of  0.5  degree 
per  minute  and  the  volume  to  be  increasing  at  the  rate  of  0.2  cubic  foot  per 
minute.     Required  the  rate  at  which  the  pressure  is  changing. 

T 
We  have  p  =  54  — , 

v 

whence  dp.  =  54     dp=_^T 

dT      v      dv  v2 


Art.  125]  TOTAL   DERIVATIVES  239 

dv  ~l  54 

Hence,  in  the  given  state,       —£-  =  —  =  3.6, 

dTJ*=T0     15 

and  &1        =-64x640=- 168.6. 

dv-iv=v0  152 

Also,  taking  the  time  t  as  the  auxiliary  variable, 
^=0.5,    and    *?=0.2. 

at  at 

Then,       #=i^+i^  =  3.6x0.5_  153.6  x  0.2  =-  28.92. 
dt      dT  dt       dvdt 

That  is,  the  pressure  is  decreasing  at  the  rate  of  28.92  lb.  per  square  foot  per 
minute. 

Ex.  2.    The  equation  z  =  2  x2  +  5  y2  represents  an  elliptic  paraboloid.     At 
the  point  x  =—  3,  y  =  1,  we  have 


dadx=-3  Jz=-3 

I?]      =10,1      =10. 


Assume  the  rate  of  change  of  x  to  be  3  units  per  second,  and  that  of  y  to  be 

^  =  3and^ 
dt  dt 


2  units  per  second  ;  that  is,  —  =  3  and  53(  =  2.    Then  the  rate  of  change  of 


z  is 

dt  dt  dt 


dz__12dx  +  l0dy=_16  unitg  per  second# 


Let  us  now  suppose  that  the  variables  x  and  y  are  not  inde- 
pendent, but  that  they  are  connected  by  some  relation  that  may 
be  expressed  explicitly  by  the  equation 

•    y  =  F{x), 

or  implicitly  by  the  relation 

<f>(x,y)  =  0. 

This  relation  restricts  the  moving  point  (x,  y)  to  a  particular 
curve  in  the  XF-plane  and  consequently  the  values  of  z  to  a 
particular  curve  in  space,  that  is,  to  the  intersection  of  the  cylin- 
der <(>(x,  y)  =  0  and  the  surface  z=f(x,  y).  Whenever  such  a 
relation  as  the  above  exists  between  x  and  y,  we  may  choose  x 
itself  as  the  variable  t,  and  formula  (a)  reduces  to  the  following : 

dz__dz_.dz_  dy  /*  x 

dx      doc      dy  dx 


240  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES     [Chap.  XIII. 

dz  dz 

In  this  formula  it  is  to  be  observed  that  —  and  —  have  very 

dx  dx       n 

different   meanings.     In  finding  the  partial  derivative   —  it  is 

ox 

assumed  that  x  alone  varies,  that  is,  that  y  is  constant.     On  the 

dz 
other  hand,   —   expresses  the  variation  of  z   due  to  a  change 
dx 

in  both  x  and  y\  for  this  reason  it  is  called  a  total  derivative. 

dz 
Likewise  in  («),  —  is  the  total  derivative  of  z  with  respect  to  t. 

Moreover,  the  value  of  —  is  definitely  determined  by  the  value 
ox 

of  the  coordinates  of  the  point  in  question ;  in  other  words,  it  is  a 

function  of  the  coordinates  only.    This  follows  from  the  fact  that 

the  variation  can  take  place  in  one  direction  only.     Functions 

like  this,  which  depend  for  their  values  solely  upon  the  coordinates 

dz 
of  the  point,  are  called  point  functions.     The  value  of  —  depends, 

dx 

however,  not  only  upon  the  coordinates  of  the  point,  but  also 

upon  the  direction  in  which   that  point   is  approached.      This 

derivative  is  therefore  not  a  point  function. 

Formulas  (a)  and  (b)  may  be  extended  to  functions  of  any 

number  of  variables.     Thus,  if 

w=/(a>,  y,  z), 

we  have  du  =  dudx  +  dudy  +  dudz.  (c) 

dt       dx  dt       dy  dt       dz  dt 

and  if  further  y  =  <f>  (x),  z  =  \f/(x), 

du  _du.dudydu  dz^  ,-,^ 

doc      dx      by  dx      dz  dx 

The  proof  is  left  as  an  exercise  for  the  student. 

Formulas  (b)  and  (d)  are  useful  in  the  differentiation  of 
somewhat  complicated  functions  of  a  single  variable.  The  fol- 
lowing example  shows  such  an  application. 

Ex.  3.     Find  — ,  where  u  =  xev~^z^  sin8  z. 
dx 


Let  y  =  Va2  —  x2  and  z  =  sin8  x  ;  then  u  =  xevz, 
ox  dy  dz 


and  ^  =  evz,     *»  =  xevz,     *»  =  X0. 


Art.  125]  TOTAL   DERIVATIVES  241 

Also  f*  = *- ,    ^  =  3sin2xcosx. 

dx  ^/a-2  _  £-2      dx 

Substituting  these  expressions  for  the  derivatives  in  (d),  we  get 

&Ol  =  &z x2e*z     +  3  x&  sin2  x  cos  x 

dx  Va2  -  x'2 

=  e^^1^  sins  x  T  i ^2        +  3  cot  xl  • 

L        Va?  -a?  J 


EXERCISES 

Find  —  in  each  of  the  following. 

1.  u  —  x2  +  y2,  and  y  =  e8inx. 

2.  u  =  arc  tan  - ,  and  y  =  ex. 

x 

3.  u  —  log  {x  -J- «/),  and  y  =  Va;2  +  a2. 

i 

4.  w  =  x2excosx. 

5.  u  =  eax  (y  —  z),  and  y  =  a  sin  x,  «  =  cos  x. 

6.  A  triangle  has  a  base  of  10  units  and  an  altitude  of  6  units.  The  base 
is  made  to  increase  at  the  rate  of  2  units,  and  the  altitude  to  decrease  at  the 
rate  of  £  unit.    At  what  rate  does  the  area  change  ? 

7.  A  point  lying  on  the  ellipsoid  —  +  ^-+  —  =  lin  the  position  x  =  3, 

36      25      49 

y  =  —  4,  moves  so  that  x  increases  at  the  rate  of  two  units  per  second,  while 
y  decreases  at  the  rate  of  three  units  per  second.  Find  the  rate  of  change 
of  z. 

8.  With  the  same  data  as  in  illustrative  Ex.  1,  suppose  the  pressure  of 
the  gas  to  be  increasing  at  the  rate  of  40  pounds  per  square  foot  per  second, 
while  the  temperature  is  falling  at  the  rate  of  1  degree  per  second.  Find  the 
rate  of  change  of  the  volume. 

9.  A  gas  has  the  equation  pv  =  I2T,  and  expands  following  the  law 
pvn  —  C,  where  n  and  C  are  constants.     Derive  expressions  for   the  total 

dT         dT 

derivatives  — -  and  —  under  these  conditions. 
dv  dp 

10.  Derive  the  same  results  by  eliminating  p  and  v  successively  between 
the  characteristic  equation  of  the  gas  and  the  equation  of  the  expansion  and 
differentiating  the  resulting  expressions. 


242  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES     [Chap.  XIII. 

126.    Total  differentials.     In  formula  (a),  — ,  — ,  and  ^  are  the 

v  J   dt    dt  dt 

quotients  of  two  differentials ;  hence  we  may  multiply  through 
by  the  differential  dt.     The  resulting  formula 

dz  =  ^dx  +  ^  dy  (e) 

due  dy 

expresses  the  differential  of  z  in  terms  of  the  differentials  of  x 
and  y.    This  formula  may  be  extended  to  any  number  of  variables. 

U  —  J  \XD   X2)  X3)    '  "  1  Xn)> 

,        du    ,      .  du    7      .   du  j      .  .    du    , 

du  =  —  dxx  -\ dx2  -\ dxz  -f  •  •  •  -\ dxn. 

dXi  dx2  dx3  dxn 

The  differential  dz  in  the  formula  (e)  is  called  the  total  differen- 
tial of  z.     The  terms  —  dx  arid  —  dy  are  called  respectively  the 
dx  dy 

partial  ^-differential  and  partial  y-differential  of  z.  These  latter  are 
frequently  denoted  by  dxz  and  dyz,  and  (e)  is  then  written  in  the 

f0rm  J  /      IN 

dz  —  dxz  +  dyz.  (e) 

Equation  (e)  therefore  expresses  symbolically  the  principle  that 
the  total  differential  of  a  function  of  several  variables  is  equal  to  the 
sum  of  the  partial  differentials. 

Ex.  1.     Let  u  =  exy2  sin  z.     Then 

^  =  exy2  sin  z,      —  =  2  exy  sin  «,      —  =  exy*  cos  *. 
dx  dy  dz 

Therefore        du  =  exy2  sin  0  die  +  2  exy  sin  z  dy  +  ex?/'2  cos  z  dz. 

The  differential  dz  of  a  function  of  two  variables  is  susceptible 
of  a  geometrical  representation  similar  to  that  for  the  differential 
of  a  function  of  a  single  variable  (Art.  46).  Let  PRQS,  Fig.  71, 
be  an  element  of  the  surface  z=f(x,  y),  and  let  a  tangent  plane 
PEGF  be  passed  through  the  point  P,  whose  coordinates  are 
(xn  Vh  zi)-  The  arbitrary  increments  of  the  independent  variables 
x  and  y  are  A#  =  dx  and  A?/  =  dy,  respectively.  From  the  figure 
we  have  evidently 

BD  =  AE  =  PA  tan  EPA  =  —  efo,  and  2)6?  =  CF  =  —  dy. 


Art.  126] 


TOTAL   DIFFERENTIALS 


243 


Hence     BG  =  BD  +  DG  =  —  dx  +  — 

ox  dy 

BG  =  dz. 


dy, 


that  is, 

The  increment  of  z  corresponding  to  the  increments  Ax,  Ay  of 
x  and  y  is  Az  =  I?Q ;  hence,  the  total  differential  dz  is  usually  dif- 
ferent from  the  incre- 
ment Az.  The  two 
symbols  dz  and  Az  rep- 
resent the  same  value 
when  the  surface  given 
by  z  =f(x,  y)  is  a  plane. 

It  is  worthy  of  notice 
that  since  dz  as  given  by 
formula  (e)  is  an  ap- 
proximation to  Az,  we 
may  use  that  formula  ° 
to  calculate  approxi- 
mately the  effect  on  a  function  z  of  small  errors  Ax  and  Ay  in 
the  observed  or  measured  values  of  the  variables  x  and  y. 

Ex.  Given  z  —  x2  —  xy.  find  approximately  the  increment  of  z  correspond- 
ing to  the  assumed  increments  Ax  =  0.02,  Ay  =  0.01,  when  x  =  8,  y  =  5. 

From  formula  (e),  dz=(2x-y)  Ax-x  A«/=0.22  - 0.08=0.14.  The  actual 
change  Az  is  found  to  be  0.1402. 


— *X 


Fig.  71. 


EXERCISES 

Differentiate  each  of  the  following  functions  by  means  of  formula  (e)  and 
verify  the  result  by  direct  differentiation. 

2.  k=r- 

x* 


1.   z  =  exy2. 

3.    z  =  sin  x  cos  y. 


4.    2  = 


«xey. 


arc  tan : 


7.   z  =  xsy*  —  x  2 1/2. 
Differentiate  each  of  the  following  functions. 
9.    u=xs+y*-2xyz.  10.    u  =  0*. 


8.   z  =  arc  cos  ^  +  arc  tan  - 


11.   u  =  sin  x  cos  y  tan  2. 


12.    m  =  arc  tan 


xy 


13.    u 


244         FUNCTIONS  OF  TWO  OR  MORE  VARIABLES     [Chap.  XIII. 

14.  From  the  perfect  gas  equation  pv  =  BT,  find  the  change  in  p  produced 
by  simultaneous  changes  of  v  and  T. 

15.  The  formula  for  the  coefficient  of  diffusion  of  a  gas  is  k  =  Cp  Tn,  where 
C  is  a  constant,  p  the  pressure,  and  T  the  absolute  temperature.  Find  the 
change  in  the  coefficient  due  to  simultaneous  changes  in  pressure  and  tem- 
perature. 

16.  Taking  the" formula  H=kssD*  for  the  horse  power  of  a  steamship, 
derive  an  approximate  expression  for  the  increase  in  horse  power  due  to  an 
increase  As  in  the  speed,  and  an  increase  AD  in  the  displacement. 

17.  If  u  =  x2y,  find  approximately  the  change  in  u  when  x  changes  from 
10  to  10.02  and  y  from  4  to  4.01.  Calculate  also  the  change  in  u  when  Ax  =  0.002 
and  Ay  =  0.001.  In  each  case  compare  these  approximate  changes  in  u  with 
the  actual  changes  as  derived  from  the  original  equation.  Interpret  the 
results  in  connection  with  Fig.  71. 

18.  Let  z  =  sin  x  cos  y.  If  x  =  22°,  y  =  37°,  find  the  change  of  z  due  to 
the  changes  Ax  =  10',  Ay  =  15'.  By  formula  (e),  calculate  the  approximate 
change  and  compare  the  result  with  the  actual  change. 

19.  Let  b  and  h  be  respectively  the  breadth  and  height  of  a  rectangle  and 
A  the  area ;  then  A  =  bh.    Interpret  geometrically  the  approximate  equation 

A  J.  =  h  Ab  +  b  Ah, 
and  show  wherein  the  approximation  lies. 

20.  Find  the  relative  error  in  the  computed  area  of  an  ellipse  due  to  errors 
in  the  measurement  of  its  semi-axes  a  and  b. 

21.  Given  a  triangle  having  sides  a,  &,  c,  and  opposite  angles  A,  B,  and 
C.  If  a  side  c  is  determined  by  measuring  two  sides  a  and  b  and  the  included 
angle  C,  show  that  the  error  Ac  is  given  approximately  by  the  equation 

Ac  =  Aa  cos  B  4-  Ab  cos  A  +  a  A Csin  B. 

127.  Differentiation  of  implicit  functions.  Suppose  that  y  is  de- 
fined implicitly  as  a  function  of  x  by  means  of  the  relation/(#,  y)  =  0. 
Since  the  function  f(x,  y)  has  by  definition  the  constant  value  zero 
for  all  values  of  x  and  y,  the  total  differential  must  also  be  zero. 
Hence,  we  have  Af  fif 

dx  dy 

Transposing  the  first  term  to  the  second  member  of  the  equation 
and  dividing  both  members  by  dx,  we  have 

df 
dp  _     3a?  ( f\ 

by 


Arts.  127,  128]    EXACT  AND  INEXACT  DIFFERENTIALS  245 

which  may  be  used  as  a  formula  for  writing  out  at  once  the  de- 
rivative —  of  an  implicit  function.     This  method  is  easier  to 
dx 

apply  than  the  earlier  one  given  in  Art.  49,  and  should  be  used  in 
practice. 

Ex.    Given  /(#,  y)  =  x2y  —  xys  =  0. 

Wehave  &=2xy-y2,   ^  =  x2-Sxy2; 

ox  dv 


therefore, 


if 

dy         dx  _      2xy  —  ys 
dx~     df~     &-Z%p 

By 

EXERCISES 

For  each  of  the  following  functions  find  the  first  derivative  by  the  method 
of  this  article. 

1.    y3  —  3  x2y  +  xy2  =  0.  2.   x3sin  y  —  y&cosx  =  0. 

3.    xyn=C.  4.    (x2  +  y2)2  +  y2  (x  -  2  a)  =  0. 

5.    x4  —  3  xy2  +  2y*  =  0.  6.    ex  sin  y  =  C. 

7.   p(v-b)n=C,  find  ^-  8.   ps-p2cosd  =  C,  find  ^. 

dv  dd 

9.  b2x2  +  a2y2  =  a2b2. 

128.  Exact  and  inexact  differentials.  In  deriving  formulas 
(a)  and  (e)  for  the  total  derivative  and  total  differential,  respec- 
tively, we  started  with  a  given  function  z=f(x,  y)  of  two  inde- 
pendent variables  x  and  y.  In  the  application  of  calculus  to 
problems  in  physics  and  mechanics,  we  frequently  meet  with 
expressions  having  forms  precisely  similar  to  (a)  or  (e)  yet 
derived  by  a  quite  different  process.  To  illustrate  this  statement 
let  us  consider  a  physical  problem,  that  of  heating  a  gas.  The  state 
of  a  gas  is  defined  by  the  absolute  temperature  T  and  the  volume 
v,  and  a  change  in  either  T  or  v  is  accompanied  by  the  absorption 
of  heat.  If  now  the  volume  v  is  kept  constant,  it  is  known  from 
experiment  that  a  change  of  temperature  AT7  is  accompanied  by 
the  absorption  of  heat  ATQ  =  cmAT,  where  cm  denotes  the  mean 
specific  heat  at  constant  volume  for  the   interval  AT.     If  we 


246  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES    [Chap.  XIII. 

assume  that  T  is  a  function  of  the  time,  say  T=<f>(t),  the  time 
rate  of  absorption  of  heat  is 

T     AM  T        AT  T  T     AT 

At  -  0   At         At  =  0      At        At  =  0        At  =  0  At 

that  is,  dIQ  =  cdT  , 

dt  dt  w 

where  c  denotes  the  instantaneous  specific  heat  at  the  beginning 
of  the  interval  At.  But  the  specific  heat  c  is  the  rate  of  absorption 
of  heat  with  respect  to  the  temperature  when  v  is  constant ;  that 

is,  c  =  — ^-     (See  Art.  19.)     Hence  we  may  write 

d^Q^dQdT  (2) 

dt       dT  dt'  K  } 

A  similar  course  of  reasoning  leads  to  the  result 

dvQ  _dQdv^  /nx 

dt  ~  dv  dt'  ^  } 

If  T  and  v  change  simultaneously,  the  total  rate  of  absorption  of 

heat  is  therefore 

dQ^d^dT^  ,dQdo.  (4x 

dt      dT  dt       dv  dt'  W 

Multiplying  through  by  dt,  we  obtain 

dQ-QdT+Wdv.  (5) 

Finally,  replacing  -^  by  c  and  —^  by  I,  the  so-called  latent  heat  of 

expansion  at  constant  temperature,  we  have 

dQ  =  cdT+ldv.  (6) 

It  will  be  noted  that  (4)  and  (5)  are  in  form  similar  to  equations 
(a)  and  (e)  but  that  they  are  derived  from  observed  relations 
between  the  increments  AT7,'  Av,  ATQ,  AVQ  and  the  coefficients  I 
and  c,  and  not  by  the  differentiation  of  a  previously  existing 
functional  relation  between  Q,  T,  and  v. 


Art.  128]         EXACT  AND   INEXACT   DIFFERENTIALS  247 

As  a  second  example,  consider  the  work  W  of  moving  a  particle 
in  a  plane,  say  the  X  F-plane.     It  is  shown  in  mechanics  that 

dW=Xdx  +  Ydy,  (7) 

where  X  and  Y  denote  respectively  the  X-  and  Y-  components  of 

the  force  acting  on  the  particle.     Therefore  (see  Ex.  15,  p.  67), 

dW  dW 

X= and  Y  = ,  and  (7)  takes  the  form 

dx  dy  w 

dW^dx+^dy.  •  (8) 

dx  dy     *  K  } 

Again  (8)  is  not  derived  by  differentiating  a  function  W=f(x,  y)  ; 
it  is  deduced  from  the  laws  of  mechanics  and  the  question  of  a 
functional  relation  between  W  and  the  coordinates  x  and  y  does 
not  enter  into  consideration  in  this  deduction. 

When  we  have  given  an  expression  like  c  dT-\- 1  dv  or  Xdx  +  Ydy 
the  question  arises  :  Can  this  expression  be  produced  by  the  differ- 
entiation of  some  function  of  the  variables  involved ;  for  example, 
can  we  find  any  function  as  Q  =  <f>(T,  v)  that  upon  differentiation 
will  produce  (6)  or  any  function  as  W=if/(x,  y)  that  will  likewise 
produce  (7)  ?  To  state  the  question  more  generally,  if  M  and  N 
are  any  arbitrarily  chosen  functions  of  x  and  y,  does  a  function 
of  the  independent  variables  (x,  y)  exist  that  will  upon  differen- 
tiation produce  Mdx+  Ndy?  Slight  consideration  will  show 
that  such  a  function  usually  does  not  exist,  and  if  it  does  exist, 
the  coefficients  M  and  N  must  satisfy  a  certain  condition.  Let  it 
be  assumed  that  there  is  such  a  function,  say  z  =  <f>(x,  y).  Then 
by  differentiation  we  obtain 

dz  =  —  dx  +  —  dy.  (9) 

dx  dy   y  w 

If  therefore  the  differentiation  of  the  given  function  produces 
M dx  -h  Ndy,  we  must  have 

Mte    N=dz  (10 

dx  dy 

that  is,  M  and  N  must  be  the  partial  derivatives  of  the  function  z 
with  respect  to  x  and  y,  respectively.  From  Art.  124,  we  have, 
with  proper  restrictions  regarding  continuity, 

d2z    =    dh     Qr   ±fdz\=d_fdz\ 
dydx      dxdy  dy\dxj      dy\dyj 


248  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES    [Chap.  XIII. 

Hence,  if  M  =  — -  and  N=  — ,  we  have 
ox  dy 

dM=dN  ,1V) 

dy       dx 

as  the  necessary  condition  that  Mdx  +  Ndy  may  be  produced  by 
the  differentiation  of  a  function  <f>(x,  y).  It  may  also  be  shown 
that  this  condition  is  sufficient.  * 

If  the  condition  (11)  is  satisfied,  M  dx-\-  N  dy  is  called  an  exact 
differential;  if  the  condition  is  not  satisfied,  Mdx  +  Ndy  is  called 
an  inexact  differential. 

Ex.  1.  Given  M  dx  +  iV  dy  =  ^  dx  +  log  x  dy. 

x 

KereM=y,   N=  log  a,  $M  =  1 ,  2^=1.     The  condition  imposed  by 

x  dy      x      dx      x 

(11)  is  satisfied  and  the  differential  is  exact.  It  is  easily  seen  that  the  func- 
tion is  y  log  x. 

Ex.  2.   Given  M dx  +  Ndy  =  x2y  dx  —  2xy  dx. 

In  this  case  we   have  ^M=A  (^y)  =  x2  and  —  =  —  ( -  2  xy)  =  -  2  y. 
dy      dy  dx      dx 

The  given  differential  is  therefore  inexact,  and  no  function  of  x  and  y  exists, 
the  differentiation  of  which  will  produce  this  differential. 

The  essential  difference  between  exact  and  inexact  differentials 
appears  more  clearly  when  integration  is  attempted.  If  the 
differential  Mdx  +  Ndy  is  exact,  it  is  then  the  total  differen- 
tial of  some  function  z  =  <f>(x,  y)  of  the  independent  variables  x 
and  y.     We  have  therefore 

f(Mdx  +  Ndy)  =Jdz=<f>(x,  y)+  C.  (12) 

If  we  assign  limits  of  integration,  as  (x1}  y^  and  (x2,  y2),  we  have 
(Mdx+  N  dy)  =  <f>(x2i  y2)  -  <j>  (xly  yx) .  (13) 


£ 


The  value  of  the  integral  therefore  depends  only  upon  the  initial 
and  final  values  of  the  variables.  If  we  represent  (a?,,  yx)  and  (x2,  y2) 
by  points  in  the  XF-plane,  we  say  that  the  integral  depends  on 


See  First  Course,  Art.  160. 


Art.  128]        EXACT  AND   INEXACT  DIFFERENTIALS 


249 


the  end  points  only  and  not  at  all  upon  the  path  by  which  the 
variable  point  moves  from  one  to  the  other.  The  integral  is  thus 
a,  point  function  of  the  coordinates  x,  y. 

Ex.  3.    Given  the  differential  y  dx  +  x  dy. 

This  differential  satisfies  the  condition  (11)  and  is  therefore  exact ;  and 
since  by  inspection  y  dx  -\-xdy  =  d(xy),  we  have 


Jxv  y1 


dx  +  x  dy) 


I        d(xy) 


mi-2  ~  ^12/1- 


This  integral  is  represented  geometrically  by  the 
shaded  area,  Fig.  72,  and  is  evidently  independent    y 
of  the  path  p  between  the  point  (cci,  y{)  and 
(z2,  y2). 

The  integration  of  an  exact  differential 
may  be  effected    by   the   following   rule,    o 
which   is   sufficient   for  most   cases  that 
arise  in  practice. 

Integrate  Mdx  considering  y  as  a  constant,  then  integrate  the  terms 
in  Ndy  that  do  not  contain  x,  and  take  the  sum  of  the  two  integrals. 

'  Ex.  4.    Given  dz  =  (3  x2  +  2  y2)dx  +  (4  xy  -  9  y2)dy. 

Since    -^-  (3  x2  +  2  y2)  =  JL  (4  xy  -  9  y2)  =  4ty,   the   differential    is    exact. 
dy  dx 

Integrating  M  dx  with  y  constant,  we  have 

f  Mdx  =  ( (3  x2  +  2  y2)dx  =  x*  +  2  xy2. 


The  part  of  N  that  does  not  contain  x  is  —  9  y2,  and 
.  Therefore  the  integral  is  z  =  xz  +  2  xy2  —  3  ys  +  C. 


j*  -9y2dy 


Sy3 


If  Mdx  +  Ndy  is  an  inexact  differential,  no  function  <f>  (x,  y)  can 
be  found  the  differentiation  of  which  will  produce  this  differential. 

Consequently  the  integral    |       2(Mdx  +  Ndy)  cannot  be  expressed 

as  the  difference  cf>(x2,  y2)—  <f>(x1,  yx),  and  to  arrive  at  any  definite 
result  we  must  assume  a  relation  between  x  and  y,  as  y  =  F(x). 
From  this  relation  we  can  obtain  an  expression  for  the  derivative 

-f-;  then  by  means  of  the  identity, 
dx 


Mdx  +  Ndy  =  (m+  N  c^)dx, 
\  dx) 


250  FUNCTIONS  OF  TWO  OR  MORE  VARIABLES    [Chap.  XIII 

we  can  express  the  integrand  in  terms  of  x  and  integrate  in  the 
usual  manner.     The  following  example  illustrates  this  process. 

Ex.  5.     Investigate  the  integral,   \      (y2  dx  —  xdy). 

Jo,  0 

Since  — —  =  2y,.— —  =  —  1,  the  differential  is  inexact  and  some  relation 
dy  dx 

between  y  and  x  must  be  assumed.     (1)  Let  the  relation  be  y  =  2  x,  which 

is  the  equation  of  a  line  passing  through  the  given  end  points  (0,  0)  and  (1,  2). 

From  this  relation  we  have  ^  =  2, 
dx 

whence  y2  dx  -  x  dy  =  (y2  -  x  Sft  dx  =  (4  x2  -  2  x)  dx, 

and  the  integral  becomes  (  (4  x2  —  2x)dx  =  \.  (2)  Let  the  assumed  rela- 
tion be  y2  =  4  x,  which  is  also  satisfied  by  the  end  points  (0,  0)  and  (1,2); 
then  dx  —  \y  dy,  and  the  integral  becomes  \  "(|  y*  —  $  y2)  dy  =  l\,  (3)  If 
the  relation  is  y  =  2  x2,  we  have  dy  =  4x dx,  and  the  integral  becomes 

f  4  (x*-x2)dx  =-TV 

It  appears  therefore  that  the  integral  of  an  inexact  differential 
is  not  determined  by  the  initial  and  final  values  of  the  variables, 
but  requires  in  addition  a  relation  between  those  variables,  that 
is,  a  path  between  the  end  points.  If  dz  =  M dx  +  Ndy  is  an  inex- 
act differential,  dz  has  therefore  no  definite  significance  as  a  total 
differential  so  long  as  x  and  y  remain  independent,  and  assumes 
definiteness  only  when  a  relation  between  x  and  y  is  furnished. 
Furthermore,  z  is  not  a  point  function  of  x  and  y. 

Returning  to  the  differentials  dQ  and  d  W,  equations  (6)  and  (7),  it' 
is  known  from  physical  considerations  that  dQ  is  inexact,  therefore 
Q  is  not  a  point  function  of  T  and  v,  and  a  relation  between  T 
and  v  must  be  established  before  the  heat  absorbed  by  the  gas 
during  a  change  of  state  can  be  determined.  If  the  force  acting 
on  a  moving  particle  is  the  force  of  gravity  alone,  or  if  it  is  a 
function  of  the  distance  of  the  particle  from  a  fixed  point,  then 
it  is  found  that  the  force  components  in  (7)  satisfy  the  relation 

r)  "ST      c\~Y 

-r—  =  -r—  •  In  this  case  d  W=  X  dx  +  Ydy  is  an  exact  differential 
dy       ox 

and  the  work  W  depends  only  upon  the  end  points  (xlf  yr)  and 
(x2,  y.2).     If  frictional  forces  are  taken  into  account,  d  Wis  not  exact 


Art.  128]        EXACT   AND   INEXACT   DIFFERENTIALS  251 

and  W  depends  upon  the  path  between  the  initial  and  final 
positions. 

EXERCISES 

Determine  which  of  the  following  differentials  are  exact,  and  for  such  as 
are  exact  find  the  functions  that  produce  them : 

1.    ex  sin  y  dx  +  ex  cos  y  dy.  2.    vn  dp  +  npv11-1  dv. 

3.   x2y>dx  +  bx*ydy.  4.    t  dx +  x  log  xdy 

5.    (x2  —  y)  dx  —  x  dy.  x 

6.  O2  -  2  xy  —  y)  dx  -  (x2  —  2  xy  +  x)  dy. 

7.  (x2  —  axy  +  y2)  dx  +  (y2  +  xy  —  ax2)  dy. 

8.  (sin  y  —  exy)  dx  +  (x  cos  y  —  ex)  dy. 

9.    Show  a  geometrical  interpretation  of  the  differential  y  dx  —  x  dy,  and 

from  purely  geometrical  considerations  show  that  the  integral  (  2'  2(ydx  —  xdy) 
must  depend  upon  the  path.  **  Vl 

10.  For  a  gas  that  follows  the  law  pv  =  BT,  we  have  I  =  p,  whence 

dQ  =  cdT  +  pdv.  Show  that  while  dQ  is  not  an  exact  differential,  — &  is  an 
exact  differential. 

/»2,  2 

11.  Find  the  value  of    the  integral   \       (ydx  —  x  dy)  for  the  following 

paths  between  (0,  0)  and  (2,  2)  :  (a)  A  path  made  up  of  the  straight  line 
joining  (0,  0)  to  (2,  0)  and  the  straight  line  joining  (2,  0)  to  (2,  2).  (&)  A 
path  made  up  of  the  F-axis  from  (0,  0)  to  (0,  2)  and  the  line  joining  (0,  2)  to 
(2,  2).     (c)  The  curve  y  =  x3  -  3  x. 

12.  Find  the  value  of  the  integral   \       (2  xy  dx+x2dy).     Show  by  choos- 

«/o,  o 

ing  two  or  more  paths  that  the  integral  is  independent  of  the  path. 

13.  If  IF  denote  the  work  done  by  an  expanding  gas,  show  that  dW  =  p  dv, 
that  dW  is  an  inexact  differential,  and  that  consequently  IF  depends  upon  the 
relations  of  p  to  v  during  the  expansion.     See  Art.  113. 

MISCELLANEOUS  EXERCISES 

Find  ^,   ^,   ^  for  each  of  the  following. 
dx     dy     dz 

1.    u  =  x2?/305.  2.    u  =  xse2v  cos  z. 

3.    u  =  (x3  +  -f  +  zrf.  4.    u  =  arc  tan  ^J^. 

z 

Find  Dxy  for  the  following  functions. 

5.   x*y2  -  4  xy*  +  3  y*  =  0.  6.    —  -  £-  -  1  =  0. 

a2      b2 


252        FUNCTIONS  OF  TWO  OR  MORE  VARIABLES      [Chap.  XIII. 


7.    ax2  +  by2  +  2  hxy  +  2ex  +  2fy  +  g  =  Q.        8. 

Find  —  in  the  following  cases. 
dx 


9.    u  =  Vx2  +  y2,   y  =  arc  tan  x. 
10.    u  =  xHx  arc  sin  x. 


vx3  —  y9 


=  Cy. 


11.  u  =  arc  tan  *■ ,    ?/  =  e_x,   3  =  cos  x. 

z 

12.  w  =  eax(y  —  z)1   y  =  a  sin  x,  at  =  cos  x. 

Form  -^-,    -^-  ,   and  2J*  for  each  of  the  following, 
dx  6y     dx2  dy  cty3 

13.  w  =  x3  -  5  x2y  +  3  ?/4.         14.    u  =  ex  cos  y.        15.   u  =  sin  x2  +  cos  xy. 

16.    Show  that    the    relation   <tX+§?K=o    holds    for    the    following 
functions.  $&       °V 


(a)    F=log(x2  +  y2);         (6)    F=arctan^. 

x 


17.    If  F  = 


—  — ,  show  that 1 1 =  0. 

Vx2  +  y1  +  z2  dx2       d?/2       dz2 

18.    The  potential  Fat  a  point  P,  in  a  straight 
line  due  to  the  attraction  of  a  mass  at  an  external 


point  C,  is 


V  =  k  log 


y  +  vg  +  y2 


Find^T  and  *F 


Fig.  73.  3x  ^ 

19.    Find  from  van  der  Waals'  equation 

the  partial  derivatives  — ^  and  -^.     Give  physical  interpretations  to  these 


derivatives. 


dT 


On 


7)2  _|_  c2  _  ^2 

20.  In  an  oblique  triangle  we  have  the  relation  cos  A  =  — — ,  where 

2  be 

a,  b,  c  denote  the  sides,  and  A  the  angle  opposite  side  a.    Find  the  rate  of 
change  of  A  with  respect  to  side  c,  keeping  sides  a  and  b  constant. 

21.  At  the  point  x  =  2,  y  =  —  3,  on  the  ellipsoid  —  +  #_  4.  ~L  =  1,  find 

/)  /)  9       25      16 

the  values  of  —  and  — .      Draw  a  figure  and  interpret  the  results  geo- 
dx         dy 

metrically. 


Art.  128]  MISCELLANEOUS   EXERCISES  253 

22.  If  z  is  a  homogeneous  function  of  x  and  y  of  degree  n,  prove  Euler^s 
theorem,  viz.  : 

ox        ay 

23.  Verify  Euler's  theorem  for  the  following  functions  : 

(a)  z  =  (x2  +  y2)  Vxy .  (6)  z  =  x3  -  3  xy2  +  2  y3. 

(c)   ^  =  arc  tan  - .  (d)  z  =  arc  cos  -  +  arc  tan  ^ . 

V  y  x 

24.  If  z  =  /(y  +  ax),  show  that 

(a)   |?=a^;     (&)<^  =  a2^. 
dx        dy  dx2         dy2 

25.  From  the  transformation  equations  x  —  p  cos  6,  y  =  p  sin  0,  derive  the 
relation 

xdy  —  y  dx  =  p2  dd. 

26.  A  point  moves  on  the  surface  z  =  x2  +  2  y2  in  a  vertical  plane  which 
includes  the  Z-axis  and  makes  an  angle  of  30°  with  the  XZ-plane.  If  the 
X-component  of  the  point's  velocity  is  3  units  when  x  =  2,  find  the  Y-  and 
Z-components  of  the  velocity. 

27.  Derive  an  approximate  value  for  the  error  in  computing  the  volume 
V  of  a  cylinder  from  the  measured  height  h  and  base  radius  r.  Find  also  an 
expression  for  the  relative  error. 

28.  Find  a  function  u  of  x  and  y  that  satisfies  the  relation 

<«)   7Tir  =  x2  +  y2>     W   T^  =  ^3cosy. 
dx  ay  dx  dy 

29.  Given  w  =  /(x,  y)  and  x  =  />  cos  0,  y  =  p  sin  0  ;  show  that 

5?*  ,      du        du 
x h  y —  =  p  — . 

dx         dy        dp 

30.  Integrate  the  following  differentials. 

(a)  2  xy  dx  +  (x2  —  y2)  dy ;     (6)  2  x  arc  tan  y  dx  +         ^  dy. 

1  +  y2 

31.  The  heat  content  i  and  entropy  s  respectively,  of  superheated  steam  are 
obtained  by  integrating  the  following  equations,  in  which  T  and  p  are  the 
independent  variables. 

di  =  (a  +  pT)dT+Amn<ini-l)p^l  +  ^p\^i-Am(^1)(l+ap)dp, 

^  =  (|+^r+^n(n  +  l)p(l  +  |p)^-^^-^/l+«i>)^ 
Find  expressions  for  i  and  s. 


CHAPTER   XIV 
MULTIPLE   INTEGRALS.     APPLICATIONS 

129.  Multiple  integrals.  In  previous  chapters  we  considered 
successive  differentiation  and  successive  integration  of  functions 
of  a  single  variable  and  the  successive  partial  differentiation  of 
functions  of  two  or  more  variables.  We  now  take  up  the  problem 
of  successive  integration  of  functions  of  several  variables. 

Suppose,  for  example,  that  we  have  given  a  function  f(x,  y,  z) 
of  three  independent  variables.     We  may  write 

J  f(p>  V,  2)  dz  =  F(x,  y,  z),  (1) 

§F{x,y,z)dy  =  Fl(x,y,z),  (2) 

JF,{x,  y,  z)  dx  =  F2(x,  y,  z),  (3) 

where  in  (1)  the  integration  is  taken  with  respect  to  z,  that  is,  as 
if  y  and  x  were  constants.  Likewise  in  (2)  it  is  taken  with 
respect  to  y,  and  in  (3)  with  respect  to  x.  Substituting  in  (3)  the 
value  of  Fx(x,  y,  z)  from  (2),  we  have 

F2(x,  y,  z)  =  JjJ>(z,  y,  z)  dy\  dx,  (4) 

and  by  the  use  of  (1)  this  becomes 

F2(x,  y,  z)  =  H  jT  ff(x,  y,  z)  dz~]dy  \  dx.  (5) 

The  expression  (5)  may  be  written  more  compactly  as  follows : 

F2(x,  y,  z)  =  ffff(x>  y> z)  dz  dv dx-  (6) 

254 


Art.  129]  MULTIPLE   INTEGRALS  255 

It  is  to  be  understood  that  the  integral  is  to  be  taken  first  with 
respect  to  z,  then  y,  and  finally  x* 

An  expression  like  the  second  member  of  (6),  which  indicates 
the  result  of  several  successive  integrations,  is  called  a  multiple 
integral.  If  two  integrations  are  involved,  the  integral  is  called  a 
double  integral;  if  three,  a  triple  integral;  etc. 

The  evaluation  of  an  indefinite  multiple  integral  differs  from 
that  of  an  indefinite  single  integral  in  one  essential  feature; 
namely,  the  form  of  the  constant  of  integration.  An  example 
will  illustrate  this  point. 

Ex.  Given      i  \  e^y2  dy  dx ;    find  a  function  u  of  x  and  y  such  that 

d2u   =  g2    2 

dxdy 
Integrating  first  with  respect  to  y  regarding  a;  as  a  constant,  we  obtain 

"    —  =  -e2xyB  +  constant  of  integration. 
dx     3 

Since  x  was  considered  as  constant  during  the  integration,  the  constant  of 
integration  may  depend  upon  x.  To  make  this  more  clear,  we  may  observe 
that  differentiating  either  j  2    3      „ 

or  \e^-h<f>(x), 

with  respect  to  y  gives  the  same  result,  e^y2.  Hence,  we  assume  the  more 
general  case  and  write  « 

as =W +#(«). 

ox     o 

where  <f>(x)  is  an  arbitrary  function  of  x.  As  a  special  case,  <f>(x)  may  of 
course  be  a  constant  C.     Integrating  this  result,  keeping  y  constant,  we  obtain 


u  =  }  e2xy*  +  f  <t>(x)  dx  +  ^(y). 


Here  again,  since  y  was  considered  constant  during  the  integration,  the  inte- 
gration constant  must  be  taken  as  an  arbitrary  function  of  y. 


*  Books  on  the  calculus  differ  in  the  manner  of  indicating  the  order  in 
which  the  integrals  are  to  be  taken.  Some  authors  write  the  differential  last 
which  is  to  be  taken  in  the  first  integration,  etc.,  that  is,  the  order  of  the 
differentials  in  equation  (6)  is  exactly  reversed.  The  above  notation  is 
adopted  in  this  book  because  it  shows  best  the  manner  in  which  the  integra- 
tion has  arisen.  In  other  works,  the  context  will  usually  indicate  to  the 
reader  the  notation  employed. 


256  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

By  assigning  limits  of  integration  to  each  variable,  we  arrive  at 
the  notion  of  a  definite  multiple  integral.     Thus  the  integral 


nVy  dV  dx 


indicates  that  e^y2  is  to  be  integrated  first  between  the  limits  a 
and*6  with  respect  to  y,  and  the  result  thus  found  is  to  be  inte- 
grated with  respect  to  x  between  the  limits  0  and  1. 

Since  x  is  considered  as  constant  in  the  integration  with  respect 
to  y,  the  y-limits,  a  and  b,  may  be  functions  of  x.  Similarly  in  a 
definite  triple  integral,  taken  first  with  respect  to  z,  then  y,  and 
finally  x,  the  z-limits  may  be  functions  of  both  y  and  x,  and  the 
^/-limits  functions  of  x. 

V 

Ex.     Evaluate  \       I      I  r2sin0  dr  ddd<p. 

The  first  integration  with  respect  to  r  gives 

i        c  2*  C  2 

-  a3  \      \     cos3  e  sin  6  dd  dd>. 

3     Jo     Jo 

Integrating  with  respect  to  0,  we  obtain 

12     >        r      6 
It  should  be  observed  that  the  upper  limit  of  the  r-integral  is  a  function  of  0. 

EXERCISES 

Evaluate  the  following  integrals. 
1.     f  \x2y  dy  dx.  2.     I  j  ex  sin  y  dy  dx. 

IT 

c«c°«-«»°\*nnedrde.        6.  re-      pdf>de. 

Jo      Jo  Jo        Ja  (1-008  0) 

t*  i    C  2  a  cos  0 

7.     I      I  ps  cos*  0  dp  d8. 


§oa$QC*^^-y\z*  +  y*)dxdydz. 


Art.  130]      PLANE   AREAS   BY   DOUBLE   INTEGRATION 


257 


11. 


JO      Jftt 

Jo    Jo 


2  gy  dy  dx. 


10 


x2y  dy  dx. 


•  n' 

Jo    Jo 


P  dd  dp. 


xn  dy  dx. 


130.    Plane  areas  by  double  integration,  rectangular  coordinates. 

We  have  seen  (Art.  100)  that  the  area  bounded  by  a  curve,  the 
axis  of  x,  and  two  or- 
dinates  is  given  by  an 
integral  of  the  form 


j   f(x)dx. 


When  the  area  is 
bounded  by  two  curves, 
as  in  Fig.  74,  the  area 
is  given,  not  by  a  single 
integral,  but  by  the 
difference  of  the  two  ^ 
integrals 


*   X 


Fig.  74. 


fbf(x)dx,  fhF(x)dx. 


The  same  result  may  be  accomplished  by  successive  integration. 
Consider  the  element  of  area  Ay  Ax.  If  we  sum  up  these  elements 
with  respect  to  y,  we  shall  have  the  area  of  the  strip  PQ,  and 
then  by  summing  up  the  strips  between  the  limits  a  and  b  we  have 


*     « 


X  \X  Ax] Ay = 2  X  Ax  Ay- 

a     L    P  J  a       P 


Upon  passing  to  the  limits  first  as  Ay  =  0  and  then  as  Ax  =  0,  we 
have  the  required  area  given  by  the  double  integral 


A-  I      f        doc  dy, 

J  a   JFyX) 


(i) 


which  leads  to  the  same  expression  for  the  given  area  as  was 
obtained  in  Art.  100. 

We  might  equally  well  have  summed  up  the  elements  of  area 


258 


MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 


in  the  reverse  order,  namely,  first  with  respect  to  x  and  then  with 
respect  to  y.     In  this  case,  we  should  have  obtained 


(2) 


where  <f>(y)  and  ij/(y)  are  the  inverse  functions  of  F(x),  /(as), 
respectively. 

Ex.  1.  Find  the  area  between  the  circle  x2  +  y2  =  a2  and  the  line  x  +  y  =  a. 
See  Fig  75. 

The  coordinates  of  the  points  of  intersection  are  (0,  a)  and  (a,  0).  The 
lower  limit  of  the  y-integral  is  the  value  of  y  found  from  the  equation 
x  +  y  =  a,  namely,  \y  =  a  —  x  ;  and  the  upper  limit  is  the  value  of  y  deter- 
mined from  the  equation  x2  +  y2  =  a2,  namely, 
y  =  Va2  —  x2.    Hence  we  have 

»*/Zs   .... 

dx 


A=\    i    a       dy< 

J0    Ja-x  * 

—  (       Va2  —  x2—(a  —  x)  \dx 

=  [-  Va2  -  x2  +  —  arc  sin  -  -  ax  +  —  1 
L.2  2  a  2  Jo 


Fig.  75. 


In  some  cases  the  abscissas  of  the  points  of  intersection  of  the 
two  curves  may  not  give  the  proper  limits  of  integration ;  in  such 
cases  it  is  necessary  to  divide  the  area  into 
two  or  more  parts. 

Ex.  2.  Find  the  area  bounded  by  the  curves 
x2  +  y2  =  25,  y2  =  *£x,  and  y  =  ft  x2.    See  Fig.  76. 

The  circle  and  the  first  parabola  intersect  at  the 
point  A  whose  coordinates  are  (3,  4),  and  the 
second  parabola  intersects  at  the  point  C,  whose 
coordinates  are  (4,  3).  From  O  to  A,  the  equation 
y2  =  Y ■  *  gives  the  Upper  y-limit,  but  from  A  to  C, 
this  limit  is  determined  from  the  equation  of  the 

circle,  x2  +  y2  =  25.    The  equation  y  =  ft  x2  of  the  lower  curve  OC  gives  the 
lower  y-limit  throughout.     Hence  we  have 


CC     ~8dydx  + 
Jo  *)&** 


dydx  =  7.55. 


Art.  131]      PLANE   AREAS   BY   DOUBLE    INTEGRATION  259 

EXERCISES 

1.  Find  by  double  integration  the  area  of  a  parallelogram  with  one  side 
in  the  .X-axis. 

2.  Find  by  double  integration  the  area  of  a  right  triangle  with  a  short 
side  in  the  X-axis. 

3.  Find  by  double  integration  the  area  between  the  parabolas  y2  =  8  x 
and  x2  =  8  y. 

4.  Find  the  area  between  the  circle  x2  +  y2  =  a2  and  the  line  y  =  b,  b  <  a. 

5.  Find  the   area  between  the  circle  x2  +  y2  =  2ax  and  the  parabola 
y2  =  ax. 

6.  Find  the  area  between  the  curve  y  =  xs  and  the  F-axis  from  the  origin 
to  y  =  8. 

7.  Find  the  area  bounded  by  the  parabolas  y2  =  -1/-  x,  y2  =  f  x  and  the 
circle  x2  +  y2  =  25.     Consider  the  first  quadrant  only. 

8.  Find  the  area  between  the  circle  x2  +  y2  =  25  and  the  line  x  +  y  =  7. 

9.  Find  the  area  bounded  by  the  curves  x2  +  y2  =  169,  y2  =  ^ff  xs,  and 
5  x—  12  y  =  0,  in  first  quadrant. 

10.    Find  the  area  of  the  ellipse  —  +  %-  =  1. 

a2     &2 

131.  Plane  areas  by  double  integration,  polar  coordinates.  When 
it  is  desired  to  find  the  area  of  a  surface  bounded  by  two  curves 
given  in  polar  coordinates,  we  may  em- 
ploy the  method  of  Art.  105,  or  we  may 
find  the  desired  area  by  double  integra- 
tion. The  latter  method  is  introduced 
here  as  a  simple  exercise  in  the  use  of 
double  integration. 

Let  the  polar  element  of  area  be  ABCD,  Fig.  "7,  bounded  by  the 
two  radii  0(7,  OB,  and  two  circular  arcs  having  their  common 
center  at  O.  Let  the  polar  coordinates  of  B  be  (p,  9).  From  ele- 
mentary geometry,  we  have 

sector  AOB  =  ip2A6,  (1) 

sector  DOC  =  Up  +  Ap)2  A0.  (2) 

Hence,  A.4  =  ABCD  =  \  (p  +  Ap)  *  A0  -  \  p2  A0 

=  (p  +  iAp)A6kp.  (3) 


260 


MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.    XIV. 


Using  this  polar  element,  we  may  find  the  area  between  two 
curves  p  =  F(0),  p  =/(0),  Fig.  78,  as  follows :  First  keep  A0  con- 
stant and  sum  the  elements  of  area  with  respect  to  p.     The  result 

of  this  summation  is  an 
area  of  the  type  PQSR, 
the  expression  for  which  is 

OP 

A6-L     X(p  +  iAp)Ap 
Ap  =  0   oq 


Fig.  78. 


=  A  0         p  dp.      (Art.  100) 
Joq 

If  now   we  sum  with  re- 
spect to  0,  we  get  the  sum 

of  the  wedge-like  slices,  and  the  limit  of  this  sum  is  the  required 

area.     We  have  therefore 

J!l  nop  /»*,  r*op 

A=L     TA0.   |      pdP=l      I     pdpdd. 

A5i0  7  Joq  Jex  Joq 

Replacing  OQ  and  OP  by  F{6)  and  /(0)  respectively,  we  obtain 
the  formula 

(4) 


A  —  I      I        pdpdQ 


The  area  included  between  the  curve  p=/(0)  and  two  radii,  as 
OA  and  OB,  is  obtained  from  (4)  by  making  F(0)  =  0. 

The  required  area  may  also  be  obtained  as  follows :  Summing 
first  with  respect  to  0,  keeping  Ap  constant,  we  obtain  a  segment 
of  a  circular  ring  of  the  type  EFOH.  A  second  summation  with 
respect  to  p  gives  the  sum  of  such  ring  segments,  the  limit  of 
which  sum  is  the  area  A.     The  resulting  formula  is 


A 


Jr*p8  ru?) 


p  dQ  dp, 


(5) 


where  <£(p)  and  \f/(p)  are  the  inverse  functions  of  F(0)  and  /(0), 
respectively. 

Ex.  Find  the  area  between  the  circle  p  =  cos  6  and  one  loop  of  the  lem- 
niscate  p2  —  cos  2  6.     Fig.  79. 

If  formula  (4)  is  used,  the  area  must  be  taken  in  two  parts.  The  upper 
limit  for  both  integrations  is  determined  from  the  equation  of  the  circle 


Art.  131] 


VOLUME   BY  TRIPLE   INTEGRATION 


261 


p  =  cos  0.     For  values  of  0  between  0  and  -,  the  lower  limit  of  the  p-integra- 

4 
tion  is  obtained  from  the  equation  of  the  lemniscate  p2  =  cos  2  0.     Since, 
however,  no  part  of  this  loop  of  the  lemniscate  lies  to  the  left  of  the  line  OA, 

/A 


Fig.  79. 

the  lower  limit  for  the  p-integration  for  values  of  0  greater  than  -  is  0. 
Hence  we  have 


Jo  Jx 


V/CO8  20 


pdpdd  +  2 


n 


cos0 


pdpdd  = 


EXERCISES 

1.  Find  by  double  integration  the  area  of  the  circle  p  =  a. 

2.  Find  the  area  between  the  cardioid  p  =  2  a(l  —  cos  0)  and  the  circle 

p  =—  a  cos  0. 

3.  Find  the  area  between  two  circles  tangent  internally  and  having  radii 
ri  and  r2  respectively.     Work  also  by  single  integration. 

4.  Find  the  entire  area  of  the  cardioid  p  =  2  a(l  —  cos  0). 

5.  Find  the  area  of  one  loop  of  the  lemniscate  p2  =  a2  cos  2  0. 

6.  Find  the  area  between  the  circle  p  =  cos  0 
and  one  loop  of  the  lemniscate  p2  =  cos  2  0  by 
the  use  of  formula  (5). 

7.  Find  the  areas  between  the  cardioid 
p  =  2  a(\  —  cos  0)  and  the  circle  p  =  2  a.  (The 
shaded  areas  OB  AD  and  BDCB,  Fig.  80.) 
Work  also  by  single  integration. 

8.  Work  Ex.  3  by  formula  (5). 


Fig.  80. 


132.  Volumes  by  triple  integration,  rectangular  coordinates. 
Consider  the  volume  bounded  by  the  coordinate  planes  and 
any  surface  whose  equation  is   z=f(x,   y)t   where  f(x,  y)  is  a 


262 


MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.    XIV. 


continuous  function.  Through  two  neighboring  points  P  and 
Q  (Fig.  81)  within  the  solid  let  planes  be  passed  parallel  to  the 
three  coordinate  planes.  The  six  planes  will  inclose  a  rectan- 
gular parallelopiped   PQ  whose  volume  is  AzAyAx\   for  if  we 


Fig.  81. 


assume  the  coordinates  of  P  and  Q  to  be  respectively  (x,  y,  z) 
and  (x-\-Ax,  y-\-Ay,  z-j-Az),  the  edges  of  the  parallelopiped  are 
Ax,  Ay,  and  Az,  respectively. 

We  first  take  the  sum  of  these  elementary  parallelopipeds  with 
edges  lying  along  the  line  S'PS.  With  x,  y,  Ax,  and  Ay  constant, 
the  limit  of  this  sum  as  Az  is  made  to  approach  zero  is  the  volume 
of  the  prism  whose  base  is  Ax  Ay  and  whose  altitude  is  S'S.  Now 
with  x  and  Ax  constant,  we  take  the  sum  of  the  prisms  lying 
between  the  planes  AMD  and  BNC.  By  an  extension  of  the 
method  of  Art.  100,  it  follows  that  as  Ay  is  made  to  approach 
zero,  this  sum  approaches  the  volume  of  the  cylindrical  slice 
AB'C'DMN.  Finally,  we  take  the  sum  of  the  slices  parallel  to 
the  YZ-plane.  As  Ax  approaches  zero,  the  volume  of  the  cylin- 
drical slice  approaches  that  of  the  actual  slice  ABCDMN.    Hence, 


Art.  132]  VOLUMES   BY   TRIPLE   INTEGRATION  263 

the  limit  of  the  sum  of  these  slices,  as  Ax  approaches  zero,  gives 
the  volume  of  the  solid.     We  have,  therefore, 

SS' 

Volume  of  prism  =  Ay  Ax    L     X  Az  ==  Ay  Ax  I       dz. 

MD 

Volume  of  cylindrical  slice  =  Ax    L     X  Ay  (  j       dz) 

Ay  =  Oo         \y°  J 

=  Ax  I        I       dz  dy. 

Jo      Jo 

OE 

Volume  of  solid  =     L     2^Ax(  (  dydz) 

Ax  =  0   o  \Jo     J  J 

j        I       dzdydx.  (1) 

o      Jo      Jo 

Hence  the  volume  is  obtained  by  a  triple  integration,  provided 
the  limits  of  integration  are  properly  chosen. 

The  first  summation,  namely,  that  with  respect  to  z,  extends 
from  zero  to  S/S',  which  is  the  value  of  the  ordinate  of  a  point  on 
the  surface.  The  upper  limit  is,  therefore,  a  variable  which  is 
given  by  z=f(x,  y),  where  x,  y  are  the  coordinates  of  the  point 
S'.  The  second  summation  is  taken  from  zero  to  MD,  and  hence 
the  limits  are  zero,  and  the  value  of  y  on  the  curve  FDE,  that  is, 
the  value  of  y  determined  from  the  equation  f(x,  y)  =  0,  it  being 
assumed  that  y  is  a  single-valued  function  of  x.  Let  this 
value  be  y  =  cf>(x).  Finally,  the  integration  which  gives  the  sum 
of  all  the  slices  between  0  and  E  has  for  its  limits  zero  and  the 
constant  OE  (=«,  say).  We  may  therefore  write  equation  (1) 
in  the  form 

F=Joio         Jo  «*»<*»*«'•  (2) 

It  will  be  observed  that  the  limits  for  the  second  and  third  in- 
tegration are  the  same  as  those  that  would  be  used  in  finding  the 
area  OEF,  i.e.  the  projection  of  the  given  solid  on  the  XF-plane, 

by  means  of  the  double  integral    J    |  dy  dx.     This  fact  suggests  a 


264  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

method  of  finding  the  required  volume  by  double  integration, 
namely,  by  the  integral 

z  dy  dx, 


Jo    «/o 


when  z  is  first  expressed  in  terms  of  x  and  y. 

If  the  given  solid  is  not  bounded  by  the  coordinate  planes,  the 
lower  limits  will  not  be  zero,  as  in  (1)  and  (2)  ;  they  can  be  readily 
determined,  however,  in  the  same  manner  as  the  upper  limits. 

EXERCISES 

1.  Determine  the  limits  of  integration  for  the  triple  integral  (   I   \dzdydx 

required  in  finding  the  volume  of  the  pyramid  bounded  by  the  coordinate 

planes  and  the  plane  -  +  ^  +  -  =  1.     Draw  the  figure. 
a     b      c 

2.  The  cone  whose  equation  is  y2  +  z2  —  c2x2  has  the  X-axis  as  its  axis. 
Determine  the  limits  of  integration  when  the  volume  of  the  cone  is  found 

from  the  triple  integral  \  \  I  dzdy  dx.    Also  when  the  volume  is  found  from 
the  triple  integral  \  \  \  dx  dy  dz.     Let  h  denote  the  altitude. 

3.  Find  the  volumes  in  Exs.  1  and  2. 

4.  Find  by  triple  integration  the  volume  of  the  ellipsoid 

£  +  £  +  £=  i. 

a2      b2      c2 

5.  Find  the  volume  inclosed  by  the  surface 

x2y2  +  c2z2  =  a?y2 
and  the  planes  y  =  0  and  y  =  c. 

6.  Show  that  the  volume  generated  by  revolving  a  plane  figure  about  the 
X-axis  may  be  found  from  the  formula 

V  =  2  w  I     i        ydydx. 

JC    J f{x) 

Compare  with  results  in  Art.  106. 

7.  Find  the  entire  volume  bounded  by  the  surface  whose  equation  is 

a*  +  tr  +  z    =  a  • 

8.  Find  the  volume  of  the  wedge  cut  from  the  cylinder  x2  +  y2  —  a2  by 
the  planes  z  —  0  and  z  =  x  tan  /S. 


Art.  133] 


VOLUMES   BY  TRIPLE   INTEGRATION 


265 


Fig.  82. 


133.  Volumes  by  triple  integration,  polar  coordinates.  Let  p,  0, 
cf>  denote  the  coordinates  of  any  point  P  within  a  solid,  where,  as 
usual,  p  is  the  distance  OP  (see 
Fig.  82),  6  is  the  angle  ZOP  which 
OP  makes  with  the  Z-axis.  and  <f>  is 
the  angle  XOP'  which  the  projec- 
tion of  OP  on  the  X  F-plane  makes 
with  the  X-axis.  Through  P  sup- 
pose three  surfaces  passed :  (1)  the 
surface  of  a  sphere  with  radius  OP 
and  O  as  a  center;  (2)  a  conical 
surface  produced  by  revolving  OP 
about  OZ  as  an  axis ;  (3)  a  plane 
surface  passed  through  OP  and  OZ. 
Now  let  a  second  spherical  surface  y 
be  passed  through  Plf  a  second  coni- 
cal surface  through  S  by  the  rotation  of  OS/Si  about  OZ,  and  a 
second  plane  surface  through  OZ  and  OQ.  The  six  surfaces 
inclose  a  solid  element  PQRSSxRxQ^  having  the  two  spherical 
surfaces  PQRS  and  PxQiR^,  the  two  conical  surfaces  PPiQxQ 
and  SS^R,  and  the  plane  surfaces  PPyS^  and  QQ^R.  By 
means  of  such  sets  of  surfaces  the  entire  solid  may  be  divided 
into  elements  of  this  type. 

Let  PP1=AP,  angle  POS=A6,  angle  P'OQ'=  angle  PHQ=  A<£. 
Then  the  sides  of  the  given  element  of  volume  are  PP'  =  Ap, 
PQ  =  PH  A<t>  =  p  sin  6  Acf>,  and  PS  =  p  A<9.  Consequently,  the 
volume  of  this  element  is  p2  sin  6  A9  Ac£  Ap,  plus  other  terms 
which  vanish  when  we  pass  to  the  limit.*  Therefore,  denoting 
the  required  volume  by  V,  we  may  write, 

V=     L        L        L     Vy,Vp2siTiOApA<p&0, 


*  The  exact  expression  for  the  volume  element  is 


p2  sin  d  A/3  Ad  A0 


i'+f) 


sin  6 


sin 


A01 


A0 
2 


V        p 


sPy 


It  will  be  seen  that  as  A0  =  0  the  factors  in  the  first  two  parentheses  approach 
1,  and  as  Ap  =  0,  the  last  factor  approaches  1. 


266  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

whence  V  =  J  J  jV2sinedPd<Me,  (1) 

each  integral  being  taken  between  the  limits  required  by  the  con- 
ditions of  the  problem  under  consideration.  This  summation  of 
the  elements  of  volume  can  be  effected  in  any  order  so  long  as 
the  given  volume  is  continuous.  The  method  of  determining  the 
limits  of  integration  is  illustrated  by  the  following  example : 

Ex.  1.     Find  the  volume  included  by  the  surface  p  =  a  sin2  6  cos  <p. 

This  volume  lies  entirely  to  the  right  of  a  tangent  plane  through  the  origin 

perpendicular  to  the  initial  line.     The  limits  for  0  are,  therefore,  —  -  and 

-,  and  the  entire  volume  will  be  included  if  6  be  given  the  limits  0  and  ir 
ss 

(see  Fig.  82).     The  limits  of  pare,  of  course,  0  and  a  sin2  0  cos  <p.     Hence 

we  have 

V=  I     \         \         p2  sin  6  dp  d<t>  dd 


=  t  f "  C      sin"  d  cos*  <pd<pdd  =  ^a 
3  Jo  J_e  r    "  315 


For  a  solid  of  revolution  formula  (1)  may  be  simplified  as  fol- 
lows:  Taking  the  Z-axis  as  the  axis  of  revolution,  the  limits  of 
the  integration  with  respect  to  <f>  are  clearly  0  and  2  w ;  hence  (1) 
becomes 

V=  2  7T  f  CP2  sin  6  dp  dd.  (2) 

The  limits  of  integration  for  p  and  6  are  precisely  those  that 
would  be  employed  in  finding  the  area  of  the  plane  figure 
revolved. 

Ex.  2.   Find  the  volume  generated  by  revolving  the  cardioid 
p  =  2a(l  —  cos0) 


about  the  initial  line. 
We  have  in  this  case 


Ctt  /*2a(l-cos0) 

F=2  7ri     I  p*sinedpd6 

=  l^i8  f  *  (l  -  cos  ey  sin  e  do  =  6i  «a 

3      J»  3 


Art.  134]  PROBLEMS   INVOLVING   SUMMATION  267 

EXERCISES 

1.  Find  by  polar  coordinates  the  volume  of  a  sphere  (a)  when  the  origin 
is  taken  at  the  center ;  (&)  when  the  origin  is  taken  at  the  end  of  a  diameter. 

2.  By  passing  to  polar  coordinates  tind  the  entire  volume  of  the  solid  in- 
eluded  by  the  surface  ^  +  y.2  +  ^  =  ^ 

3.  In  the  same  manner  find  the  volume  included  by  the  surface 

(x2  +  y2  +  z2)3  =  27  asxyz. 

4.  Find  the  volume  obtained  by  revolving  the  lemniscate  p2  =  a2  cos  2  0 
about  the  initial  line. 

5.  Find  the  volume  obtained  by  revolving  the  loop  of  the  curve 

cos  2  0 

p  =  a 

cos  0 

about  the  initial  line. 

6.  Find  the  volume  obtained  by  revolving  the  curve  p  =  2  a  cos  0  +  b 
about  the  initial  line.     First  plot  the  curve  and  determine  proper  limits  for  0. 

7.  Show  the  different  types  of  solid  elements  produced  when  the  summa- 
tion is  effected  in  the  following  orders  ; 

(a)  with  respect  to  p,  then  0,  then  <p  ; 
(6)  with  respect  to  p,  then  <p,  then  0  ; 
(c)  with  respect  to  6,  then  0,  then  p. 

134.  Additional  examples.  In  the  solution  of  problems  that 
involve  the  summation  principle  there  are  three  steps :  (1)  The 
choice  of  an  element;  (2)  the  determination  of  the  limits  of 
integration  so  that  the  whole  of  the  region  involved,  and  only 
that  region,  shall  be  included ;  (3)  the  integration.  The  beginner 
is  likely  to  encounter  more  difficulty  in  the  first  two  processes 
than  in  the  third ;  in  other  words,  the  setting  up  of  the  integral 
with  proper  limits  of  integration  is  usually  the  difficult  part  of 
the  problem.  In  attacking  such  a  problem,  therefore,  attention 
should  first  be  directed  to  the  type  of  element.  While  in  most 
cases  the  rectangular  and  polar  elements  heretofore  used  are 
most  convenient,  it  is  frequently  possible  to  choose  other  types 
that  lead  more  directly  to  the  desired  result.  It  is  generally 
preferable  to  take  the  element  so  that  only  a  single  integration 
is  necessary ;  and  frequently  the  polar  element  leads  to  a 
simpler    integration    than    the    rectangular    element.       Having 


268 


MULTIPLE   INTEGRALS.    APPLICATIONS     [Chap.  XIV. 


chosen  the  element,  care  must  be  taken  to  fix  upon  proper  limits 
of  integration.  Sometimes  it  may  be  necessary  to  trace  the  curve 
representing  the  given  function  to  determine  these  limits,  but 
usually  the  limits  are  readily  found  from  the  conditions  present. 
The  following  problems  are  chosen  to  illustrate  more  fully  the 
process  of  setting  up  the  integral.  They  should  be  carefully 
studied. 

Ex.  1.     Find  the  volume  bounded  by  the  surface 1-  -m-  =  2  z  and  the 

a       b 

plane  z  —  c.     See  Fig.  83. 

The  most  direct  method  of  solu- 
tion is  to  take  the  rectangular  vol- 
ume element  Ax  Ay  Az,  hut  this  choice 
leads  to  inconvenient  limits  of  in- 
tegration and  a  triple  integral  diffi- 
cult to  evaluate.  Slight  consideration 
of  the  conditions  shows  us  that  the 
desired  volume  can  be  obtained  by 
the  summation  of  slices  parallel  to 
the  XT-plane.  (See  Art.  107.)  Such 
slices  evidently  have  ellipses  for  their 
boundaries,  and  if  we  denote  by  x', 
y'  the  semiaxes  of  the  elliptical  cross 
section  at  a  distance  z  from  the  XY- 
plane,  the  volume  of  one  of  the  slices 
is  irx'y'Az.  From  the  parabolic  sections 

—  =  2  z  and  £-  =  2  z  obtained  by  intersecting  the  surface  with  the  XZ-  and 
a  b 

YZ-planes  respectively,  we  have  x'y'  =  2  z  Vab,  whence  the  volume  element 

is  Av  =  2irVabz  Az.     The  solid  extends  from  the  plane  z  =  0  to  the  plane 

z  =  c,  therefore  the  entire  volume  will  be  included  if  0  and  c  are  taken  as 

the  limits   of    integration.     Hence  we  have   the  integral     (    2  iry/abz  dz, 
and  performing  the  integration,  we  get  V  =  -rrc2Vab. 


Fig.  83. 


Ex.  2.  A  plane,  whose  intercepts  on  the  X-,  F-,  and  Z-axes  are  respec- 
tively a,  6,  and  c,  cuts  a  circular  cylinder  of  radius  r  standing  on  the  XT- 
plane  with  its  axis  coincident  with  the  Z-axis.  Required  the  volume  of  the 
quarter  cylinder,  Fig.  84,  intercepted  between  the  given  plane  and  the  XY- 
plane. 

In  this  problem,  again,  the  usual  volume  element  Ax  Ay  Az  may  be  chosen. 
The  conditions  are  such,  however,  that  another  type  of  element  is  better. 
To  form  this  element,  let  radii  OE  and  OF  be  drawn  in  the  base  of  the 
cylinder  and  let  6  denote  the  angle  AOF.     Between  these  radii  we  take  a 


Art.  134] 


PROBLEMS  INVOLVING  SUMMATION 


269 


polar  element  of  area  p  Ad  Ap  and  upon  this  as  a  base  erect  a  prism  meeting 
the  oblique  plane  at  Q.     The  altitude  of  the  prism  is  the  distance  PQ  between 

the  XF-plane  and  the  oblique  plane,  whose  equation  is-  +  |  +  -=l;  hence, 

CL        0        C 


if  sc,  y  are  the  coordinates  of  P,  then  PQ=z  =  c 


(*-H) 


) ,  and  the  volume 


Fig.  84. 


element  is  zp  Ad  Ap  =  c  ( 1 | )  p  Ad  Ap.  Consider  now  the  limits  of  inte- 
gration. If  we  keep  6  constant  and  vary  p  from  p  =  0  to  p  =  r,  we  get  the 
volume  of  the  wedge  OCMNFE ;  then  by  taking  the  sum  of  such  wedges 

between  the  XZ-plane  (d  =  0)  to  the  FZ-plane  Id  =  -\we  get  the  required 

volume.  Replacing  x  and  y  by  p  cos  d  and  p  sin  0,  respectively,  we  have  the 
volume  given  by  the  double  integral 

n 

Performing  the  integration,  we  have 

■>*E-S(H)]- 


270 


MULTIPLE   INTEGRALS.     APPLICATIONS       [Chap.  XIV. 


Ex.  3.     Required  the  volume  of  a  hollow  pipe  bent  as  shown  in  Fig.  85, 
the  axis  of  the  pipe  being  a  circular  arc  of  radius  B  subtending  an  angle  /3. 

Take  ri  and  r2  respectively 
as  the  inner  and  outer  radii 
of  the  cross  section. 

In  the  cross  section  of  the 
pipe  we  choose  an  element 
of  area  pAdAp,  where  p  is 
the  distance  of  the  element 
from  the  center  O  of  the 
cross  section.  Evidently 
the  distance  of  this  element 
from  an  axis  through  0'  is 
B  —  p  sin  6.  We  may  there- 
fore take  as  the  volume 
element  a  rod  of  cross  sec- 
tion p  Ad  Ap  bent  into  an 
arc  of  radius  B  —  p  sin  6  and 
subtending  an  angle  /3.  The 
length  being  P(B  —  p  sin  0) , 
the  volume  is 

p(B  -  psind)pAdAp. 
The   limits  for    p    are    ob- 
viously ri  and  r2 ;  and  one 
half  of  the  required  volume,  that  on  one  side  of  the  plane  MN,  will  be 
obtained  by  taking  —  ^  and  -  as  the  limits  for  6.     Hence  we  have 

V  =  2  f  n  C2(3  (B-p  sin  6)  pdddp  =  ttQB  (r22  -  ns). 

2 

A  still  easier  solution  is  given  by  the  theorem  of  Pappus  (Art.  137). 


EXERCISES 

1.  In  illustrative  example  2,  let  the  quarter  cylinder  be  intersected  by  a 
sphere  of  radius  c  with  its  center  at  the 

origin.  Using  the  same  type  of  volume 
element,  determine  the  limits  of  inte- 
gration and  find  the  volume  of  the  part 
of  the  quarter  cylinder  intercepted  be- 
tween the  sphere  and  the  XF-plane. 

2.  A  conoid  is  generated  by  a  line 
kept  parallel  to  a  given  plane  and  moved 
so  as  to  keep  in  contact  with  an  ellipse 
and  with   a  straight  line  AB,  Fig.  86,  Fig.  80. 


Art.  135] 


MASS.     MEAN  DENSITY 


271 


parallel  to  the  plane  of  the  ellipse.  The  semiaxes  of  the  ellipse  are  a  and  b, 
and  c  is  the  distance  of  the  line  from  the  plane  of  the  ellipse.  Find  the 
volume  of  the  conoid  (a)  by  taking  elements  parallel  to  the  base  j  (&)  by- 
taking  elements  perpendicular  to  the  line  AB. 

3.  Find  the  volume  OABCD, 
Fig.  87,  founded  by  the  three  coordi- 
nate planes  and  the  warped  surface 
generated  by  the  line  EF,  which  re- 
maining parallel  to  the  XF-plane 
slides  on  the  lines  AD  and  BC. 

If  OC  =  a,  OD  =  b,  and  OA  =  c, 
and  Az  Ay  Ax  is  taken  as  the  element, 
show  that  the  limits  of  integration 

^      ^     and 


for  z  are  0  and  c    1 


fl flL_Y 

\        b(a-x)J 


Also 


KX" 


Fig.  87. 


those  for  y  are  0  and  -  (a  —  x). 

a  y* 

choose  an  element  such  that  a  single 

integration  is  sufficient. 

4.  Show  four  different  elements  of  area  that  may  be  employed  in  finding 
the  area  of  an  ellipse.  Write  the  integrals  and  determine  the  proper  limits 
of  integration  in  each  case. 

5.  Show  that  the  volume  of  a  solid  of  revolution  may  be  found  by  means 
of  the  double  integral  2x  f  \  ydydx.  What  type  of  solid  element  leads  to 
this  integral  ?    Determine  the  proper  limits  of  integration. 

6.  Find  the  volume  of  the  part  cut  from  a  sphere  of  radius  a  by  a  cylinder 
of  radius  o,  one  element  of  which  contains  the  center  of  the  sphere. 

7.  A  right  cylinder  whose  intersection  by  the  XT-plane  gives  loop  of  the 
lemniscate  p2  =  a2  cos  2  6  is  cut  by  a  plane  that  intersects  the  XF-plane  in  the 
F-axis  and  makes  an  angle  of  45°  with  the  XF-plane.  Find  the  volume  of 
the  solid  bounded  by  the  cylindrical  surface,  the  XF-plane,  and  the  oblique 
plane. 

135.    Mass.     Mean  density,     If  m  denote  the  mass  and  Fthe 

volume  of  a  body,  the  ratio  —  is  called  the  mean  density  of  the 

body.     If  we  take  a  volume  element  AF,  inclosing  a  point  P, 

and  denote  the  mass  of  the  element  by  Am,  the  ratio  — —  is  the 

mean  density  of  the  element;  and  the  limit  of  this  ratio  as  AF 
approaches  zero  (still  including  the  point  P)  is  the  density  at  the 


272  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

point  P.  If  the  density  is  the  same  at  all  points  within  a  body* 
the  body  is  said  to  be  homogeneous;  otherwise,  it  is  said  to  be 
non-homogeneous. 

Let  the  mass  of  a  non-homogeneous  body  be  divided  into  ele- 
ments Am^  Ara2,  •••,  Amn,  whose  volumes  are  respectively  AFb 
AF2,  •••,  AFM.  Denoting  by  ylf  y2,  •••,  yn  the  mean  densities  of 
these  elements,  we  have 

Am1  =  y1AF1,  Ara2  =  y2AF2,   •••,  Amn  =  ynAFM.  (1) 

The  mass  of  the  body  is  therefore 

m  =    L      VAmk=    L      VykAVk=  CydV,  (2) 

n  =  co    i  n  =  go     i  J 

and  the  mean  density  of  the  body,  which  we  shall  denote  by  y,  is 
given  by  the  equation 

CydV 

y=™=dl (3) 

7     V         V  w 

If  the  mass  is  distributed  continuously  over  a  surface,  we  may 
replace  the  element  of  volume  AFby  an  element  of  area  AA-,  and 
if  the  mass  is  distributed  along  a  curve,  as,  for  example,  along  a 
thin  wire,  we  replace  A  V  by  As,  the  element  of  length. 

The  element  of  volume  A  V  should  be  so  chosen  as  to  lead  to  the 
simplest  integrations.  Usually  triple  integration  will  be  neces- 
sary, but  in  some  cases  the  element  may  be  taken  in  such  a  way 
that  a  double  or  even  single  integration  is  sufficient. 

Ex.  1.  Find  the  mean  density  of  a  sphere  in  which  the  density  varies  as 
the  square  of  the  distance  from  the  center. 

Since  the  distance  p  of  the  volume  element  from  the  center  determines  the 
density,  it  is  evident  that  a  polar  element  should  be  chosen.  From  the  given 
law  we  may  take  the  density  at  a  distance  p  from  the  center  as  kp2,  k  being  a 
constant ;  and  for  the  volume  element  A  V  we  take  p2  sin  d  Ap  A<p  Ad.  From 
(3)  we  have  therefore 

\     \      \    kp*  sin  d  dp  d<p  dd 

The  result  may  also  be  obtained  by  a  single  integration.  Since  the  density 
is  constant  for  all  points  at  a  distance  p  from  the  center,  we  may  choose 


Art.  186]  FIRST   MOMENTS.     CENTROIDS  273 

for  the  volume  element  a  spherical  shell  of  thickness  Ap.    We  thus  obtain 
AF=4  wp2  Ap,  whence 

47r  I    kp^dp 

The  mean  density  is  therefore  §  of  the  density  at  the  surface. 

Ex.  2.     Find  the  mass  and  mean  density  of  a  thin  plate  in  the  form  of  a 

x2      v2 
quadrant  of  the  ellipse  —  +  *-  =  1,  assuming  that  the  density  varies  as  the 

product  xy.  a 

In  this  case  we  naturally  choose  the  rectangular  area  element  A  A  =  Ax  Ay. 

The  density  may  be  noted  by  kxy,  k  being  a  constant.     Hence,  we  have 


.70  JO 


v     a'; kxydydx  =  \ka2b2\ 


and  y  =  iM*!  =  l_kab. 


EXERCISES 

1.  Find  the  mass  and  mean  density  of  a  semicircular  plate  of  radius  a, 
whose  density  varies  as  the  distance  from  the  bounding  diameter.  Take 
(a)  the  rectangular  element  of  area,  (6)  the  polar  element. 

2.  In  Ex.  1  let  the  density  vary  as  the  distance  from  the  center  ;  find  the 
mass  and  mean  density.  Take  the  element  of  area  so  that  only  a  single  inte- 
gration is  required. 

3.  Find  the  mean  density  of  a  straight  wire  of  length  I,  the  density  of 
which  varies  as  the  distance  from  one  end. 

4.  Find  the  mass  and  mean  density  of  a  hemispherical  solid,  radius  a,  the 
density  varying  as  the  distance  from  the  base. 

5.  Find  the  mass  and  mean  density  of  a  thin  plate  in  the  form  of  a  right 
triangle  in  which  the  density  varies  as  the  distance  from  one  of  the  short 
sides. 

6.  Given  a  right  circular  cone  of  height  h,  in  which  the  density  varies  as 
the  distance  from  a  plane  through  the  vertex  perpendicular  to  the  axis.  Find 
the  mean  density. 

136.  First  moments.  Centroids.  Let  a  given  geometrical  mag- 
nitude, line,  surface,  or  solid,  be  divided  in  any  convenient  way 
into  elements  —  a  line  into  elements  of  length,  a  surface  into  area 
elements,  a  solid  into  volume  elements.  Let  each  element  (As, 
AA,  or  AT)  be  multiplied  by  the  distance  of  some  chosen  point 
within  the  element  from  a  reference  line  or  plane.     The  limit  of 


274 


MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 


the  sum  of  these  products  as  the  elements  are  taken  smaller  and 
smaller  is  called  the  first  moment  of  the  geometrical  magnitude 
with  respect  to  the  given  plane  or  line.  We  shall  denote  first 
moments  by  the  symbol  M  with  appropriate  subscripts. 

fr 


m 


*-x 


Fig.  88. 


Fig.  89. 


From  the  definition  we  have  for  the  first  moment  Mx  of  a  plane 
curve  with  respect  to  the  X-axis,  Fig.  88, 

==     L     TyAs=  (yds; 


M, 


A*=0 


(i) 


and  for  the  first  moment  of  a  plane  area  with  respect  to  the  same 

axis,  Fig.  89,  ^  r 

Mx=     L     Vy&A^fydA.  (2) 

AA  =  0  J 

The  first  moment  of  a  solid  with  respect  to  one  of  the  coordinate 
planes,  say  the  Xl^plane  (see  Fig.  81),  is  given  by  the  equation 

Mxy=     L     XzAV=  CzdV.  (3) 

For  A-4  and  A  V  proper  area  or  volume  elements  are  to  be  sub- 
stituted before  integration  is  attempted.  In  some  cases  the  ele- 
ments may  be  so  chosen  that  a  single  integration  is  sufficient,  but 
in  general  double  or  triple  integration  will  be  required. 

The  first  moment  of  the  mass  of  a  solid  is  derived  from  the 
corresponding  moments  of  the  geometrical  magnitude  by  the 
introduction  of  a  density  factor.  Thus,  denoting  the  density  by 
y,  the  first  moment  with  reference  to  the  XF-plane  of  a  mass 
distributed  throughout  a  given  region  is 


Mr. 


=fyZd 


the  limits  of  integration  being  taken  so  as  to  include  the  region 
in  question. 


Art.  136]  FIRST   MOMENTS.     CENTROIDS  275 

Let  a  given  solid  be  referred  to  a  system  of  rectangular  coordi- 
nates, and  let  Mxy,  Myz,  Mzx  denote  the  first  moments  with  respect 
to  the  three  coordinate  planes.  It  is  possible  to  find  a  point  G 
whose  coordinates  x,  y,  z,  are  given  by  the  equations 


x  — 


M 


lis 


V 


CxdV 

fydv 


z  — 


V  V 


W 


V  V 

The  point  G  thus  denned  is  called  the  centroid  of  the  volume  V. 

If  we  replace  V  and  d  V  by  A  and  dA,  formulas  (4)  give  the 
centroid  of  a  plane  surface.  Likewise  the  centroid  of  a  curve 
may  be  obtained  by  using  s  and  ds,  where  s  denotes  the  length  of 
the  curve. 

If  in  (4)  we  substitute  the  mass  m  of  the  solid  for  its  volume 
V,  the  resulting  values  of  x,  y,  z  are  the  coordinates  of  the  cen- 
troid of  the  mass.*  If  the  solid  is  homogeneous,  the  constant 
density  factor  y,  which  appears  in  each  member  of  the  equation 
in  the  first  degree,  may  be  dropped,  and  in  this  case  the  centroid 
of  the  mass  and  that  of  the  volume  coincide.  If,  however,  the 
solid  is  not  homogeneous,  y  is  variable,  and  we  have 

L     2%AF=  CydV, 


m  = 

AF=0 


Myz  =  fyxdV,  etc., 

CyxdV  CyydV  CyzdV 

whence         «=^ >    V=    r     .     >    7i  =  ~^ (5) 

J    ydV  J ydV  lydV 

The  remarks  of  Art.  134  relative  to  the  choice  of  the  type  of 
element  and  the  determination  of  the  limits  of  integration  apply 
with  equal  force  here.  Very  often  the  problem  can  be  simplified 
by  care  in  selecting  the  element. 

*  The  centroid  of  a  mass  is  often  called  the  center  of  gravity  of  the  mass. 


276  MULTIPLE   INTEGRALS.    APPLICATIONS     [Chap.  XIV. 

137.  General  theorems  relating  to  centroids.  The  following 
general  theorems  are  useful  in  the  determination  of  centroids. 

Theorem  I.  Tlie  first  moment  of  a  volume  {or  mass)  with  respect 
to  a  plane  or  axis  containing  the  centroid  is  zero. 

If  the  centroidal  plane  (i.e.  plane  containing  the  centroid)  is  a 
coordinate  plane,  the  theorem  follows  at  once  from  (4)  or  (5). 
Thus,  if  the  centroid  lies  in  the  FZ-plane,  x  =  0,  whence  MVM  =  0. 
It  can  be  readily  shown  that  the  theorem  holds  for  any  centroidal 
plane ;  therefore  it  holds  for  the  intersection  of  two  such  planes, 
that  is,  a  centroidal  axis. 

Theorem  II.  A  plane  of  symmetry  of  a  solid  or  of  a  homogene- 
ous mass  is  a  centroidal  plane.  Likewise  an  axis  of  symmetry  of  a 
plane  figure  is  a  centroidal  axis. 

Take  the  plane  of  symmetry  as  a  coordinate  plane,  say  the 
XFplane.  To  an  element  A  V  above  the  plane  there  corresponds 
an  equal  element  at  the  same  distance  below  the  plane.  The 
moment  of  the  first  element  with  respect  to  the  plane  is  zx  A  V, 
that  of  the  second  element  is  —  zx  AF,  whence  the  moment  of  the 
pair  of  elements  is  zero.  Since  all  the  elements  of  the  solid  can 
be  thus  arranged  in  pairs,  the  first  moment  of  the  solid  with  re- 
spect to  the  plane  is  zero. 

Theorem  III.  If  Vlf  V2,  •••,  Vn  are  n  volumes,  and  ifxx,  x2,  x3, 
•  ••,  xn  are  the  coordinates  of  their  centroids,  then  the  x-coordinate  of 
the  centroid  of  the  system  of  volumes  is 

__Vlxl+V2x2  +  •  ••  +Vnxn  m 

•~     Fi+F.+  -  +VH    '  (1) 

A  similar  theorem  holds  for  n  masses,  mu  m2,  •••,  mn,  or  for  n 
plane  areas  in  the  same  plane.  The  numerator  of  the  second 
member  of  (1)%  is  the  moment  Mys  of  the  system  with  respect  to 
the  FZ-plane,  and  the  denominator  is  the  total  volume;  hence 
the  theorem  follows  from  (4),  Art.  136. 

Theorem  IV.     (TJieor ems  of  Pappus  and  Ouldin.) 
(a)  If  a  plane  curve  is  revolved  about  an  axis  in  its  plane,  the 
area  of  the  surface  generated  is  equal  to  the  product  of  the  length  of 
the  curve  arid  the  circumference  of  the  circle  described  by  the  centroid 
of  the  curve. 


Art.  137] 


CENTROIDS 


277 


(b)  If  a  plane  figure  is  revolved  about  an  axis  in  its  plane,  the 
volume  of  the  solid  of  revolution  generated  is  equal  to  the  product  of 
the  area  of  the  figure  and  the  circumference  of  the  circle  (or  the  length 
of  the  circular  arc)  described  by  the  centroid  of  the  area. 

From  (4),  Art.  136,  we-have  for  a  plane  curve 


/' 


ds 


or  sy 


-s 


yds, 


where  s  denotes  the  length  of  the  curve.     The  surface  generated 
by  the  revolution  of  the  curve  about  the  X-axis  is  S  —  2  ir  I  y  ds. 

(See  Art.  110).    Hence  equating  the  two  expressions  for  J  y  ds,  we 

S  =  2*ys.  (2) 


obtain 


Evidently  the  theorem  holds  equally  well  for  an  incomplete  revo- 
lution. 

To  prove  the  second  theorem,  let  the  axis  of  revolution  be  taken 
as  the  axis  of  x,  as  before ;  then  denoting  by  AA  an  element  of 
the  plane  area,  we  have  for  the  volume  generated  by  a  complete 
revolution  of  the  area, 


V=   L   S/2  7ryAA  =  27rfydA. 
AA  =  0^  J 

JydA  =  Ay; 


2  Try  A 


But 

hence, 

Ex.  1.  Find  the  coordinates  of  the  centroid 
of  a  circular  arc  of  radius  a  which  subtends  an 
angle  2  £  at  the  center,  Fig.  90. 

Take  the  line  of  symmetry  as  the  X-axis  ; 
then  y  =  0,  and  the  centroid  lies  in  this  line. 
Polar  coordinates  lead  to  the  simplest  integral ; 
hence  taking  x  =  a  cos  6,  As  =  a  Ad,  we  get, 


Mv 


£ 


2  cos  d  dd 


s  Zap 

For  a  semicircumference  /3 


_  2  «2  sin  ft  _  a  sin  /3 
2  a/3  p 


Fig.  90. 


- ,  whence  x  =  — 

2  ic 


278 


MULTIPLE    INTEGRALS.    APPLICATIONS     [Chap.  XIV. 


Ex.  2.    Find  the  coordinates  of  the  centroid  of  the  plate  described  in  illus- 
trative example  2,  Art  135. 

(  yx  dA 

From  the   defining  equation,  x  —  * 

kxy  for  y  and  dy  dx  for  dA,  j  y  dA 

b 


,  we  have,  upon  substituting 


Similarly, 


V 


Jo  Jo        kx*y  dv dx 

b, . 

fa  faVa'-~X* 

Jo  Jo        kxy  dy dx 


a«. 


Ex.  3.    Find  the  centroid  of  a  semicircle  of  radius  a. 

Taking  the  X-axis  as  the  axis  of  symmetry,  the  length  of  the  path  de- 
scribed by  the  centroid  as  the  semicircle  is  revolved  about  the  Y-axis  is 
2  irx.  The  semicircle  by  its  revolution  describes  a  sphere  whose  volume 
is  \  rraz ;  hence  by  the  second  theorem  of  Pappus, 

2  TTX  •  $  ltd2  =  $ 7m3, 


Ex.  4.   Find  the  centroid  of  a  wedge-shaped  solid,  Fig.  91,  cut  from  the 

cylinder  x'2  +  y2  =  a2  by  the  planes 


Fig.  91. 


*=0,  J+5 

b     a 


1. 


The  volume  of  the  solid  is 
|  wa2  •  2  &  =  iva2b. 
The  first  moment  with  respect  to 
the  XF-plane  is  given  by  the  in- 
tegral \  \  (zdzdy  dx.     The  limits 

of  integration,   as  determined  by 
an  inspection  of  the  solid,  are  as 

follows :     For  *,  0  and  6  (l  --\. 

for  y,  —  Va2  —  x2  and  Va2  —  x2  ;  for  x,  —  a  and  a.     Hence,  we  have 


fa     Sa*-J     Al-l) 


//-' 


f-a  (  l  ~  ~Y  ^  ~  X%  dX  =  *  ****** 


Art.  137]  CENTROIDS  279 

For  the  moment  with  respect  to  the  FZ-plane,  we  have 


ra     r  va?-x"     /»    \       a) 

^=J-J-v^J«  xdzdydx 

=  2b(a  x(l--\  Va?  -x2dx=- 


m3b 
4 


Since  the  XZ-plane  is  a  plane  of  symmetry,  y  =  0.     The  coordinates  of  the 
centroid  are  therefore 

ira2b  va2b 


EXERCISES 

Find  (a)  the  coordinates  of  the  centroids  of  the  following  plane  curves  ; 
(&)  the  coordinates  of  the  centroid  of  the  area  between  each  of  the  curves 
and  the  X-axis. 

1.  A  semicircumference. 

2.  An  arc  of  the  parabola  y2  =  4  ax  from  x  =  0  to  x  =  a. 

X  _x 

3.  An  arc  of  the  catenary  y  =  -  (ea  +  e   •)  from  x  =  0  to  x  =  a. 

4.  One  arch  of  the  cycloid  x  =  ad  —  a  sin  0,  y  =  a  —  a  cos  0. 

a*3 

5.  Find  the  centroid  of  the  area  between  the  cissoid  rj1  =  — ■ —  and   its 

asymptote  x  =  a. 

6.  Find  the  centroid  of  a  segment  of  the  parabola  y2  =  4  ax  cut  off  by  the 
double  ordinate  corresponding  to  x  =  a,  assuming  that  the  density  at  any 
point  is  proportional  to  the  distance  from  the  F-axis. 

7.  Find  the  centroid  of  a  semicircle  of  radius  a,  assuming  the  density 
proportional  to  the  distance  from  the  bounding  diameter. 

8.  Find  the  centroid  of  a  segment  of  a  homogeneous  paraboloid  of  revo- 
lution between  the  origin  and  x  =  a. 

9.  Find  the  centroid  of  a  hemisphere  whose  density  varies  as  the  dis- 
tance from  the  base. 

10.  Find  the  coordinates  of  the  centroid  of  one  loop  of  the  curve 

p  =  a  sin  2  6. 

11.  Find  the  coordinates  of  the  centroid  of  the  mass  of  one  eighth  of  a 
sphere  included  between  the  coordinate  planes,  assuming  that  the  density 
varies  as  the  distance  from  the  origin. 


\ 

^\ 

1 

I 

« 3- — >\ 

'ig.  92. 

280  MULTIPLE   INTEGRALS.    APPLICATIONS     [Chap.  XIV. 

12.  Erom  Theorem  III,  Art.  137,  show  that  if  points  P  and  Q  are  the 
centroids  of  two  volumes  Vi  and  V2  (or  areas  A\  and  A%,  or  masses  mi  and 
m2),  the  centroid  of  the  whole  system  V\  +  F2  lies  on  the  line  joining  P  and 
Q  and  divides  it  into  segments  inversely  proportional  to  V\  and  V2. 

13.  A  solid  is  composed  of  a  right  circular  cylinder  and  a  right  cone. 
See  Fig.  92.     Find  the  centroid  of  the  solid. 

14.  Three  masses,  mi  =  3,  m2  =  4,  and  m3  =  5, 
have  their  centroids  located  at  the  three  vertices 
of  an  equilateral  triangle  the  side  of  which  has 
the  length  a.  By  means  of  Theorem  III,  find 
the  position  of  the  centroid  of  the  system. 

15.  Using  the  Theorem  of  Pappus,  find  the 
centroid  of  the  semicircumference  of  a  circle  of  radius  a. 

16.  Using  the  Theorem  of  Pappus,  find  the  volume  of  the  pipe  described 
in  the  illustrative  example  3,  Art.  134. 

17.  Work  the  illustrative  example  4  of  this  article  by  taking  p  Ad  Ap  Az  as 
the  element  of  volume. 

138.  Second  moment.  Radius  of  gyration.  Let  each  of  the 
elements  into  which  a  volume  (or  area,  length,  or  mass)  is  divided 
be  multiplied  by  the  square  of  the  distance  of  some  chosen  point 
in  the  element  from  a  reference  line  or  plane.  The  limit  of  the 
sum  of  these  products  as  the  elements  are  taken  smaller  is  called 
the  second  moment  *  of  the  magnitude  in  question  with  respect  to 
the  line  or  plane  of  reference. 

Formulas  for  second  moments  may  therefore  be  derived  from 
those  for  first  moments  by  squaring  the  distance  factor.  Denot- 
ing the  second  moment  by  the  general  symbol  I,  we  have  the  fol- 
lowing formulas  corresponding  to  (1),  (2),  and  (3)  of  Art.  136. 

For  a  plane  curve,  Ix—  I  y2  ds.  (1) 

For  a  plane  area,  £=  I  y2dA.      .  (2) 

For  a  volume,  Ixy  =fz2  dV.  (3) 


*  The  second  moment  of  a  mass  is  sometimes  called  the  moment  of  inertia 
of  the  mass. 


Art.  138]  SECOND  MOMENT  281 

From  its  definition,  the  second  moment  of  a  volume  must  have 
for  its  physical  dimensions 

volume  x  (length)2 ; 

hence,  the  second  moment  of  a  volume  V  can  be  expressed  in  the 

form 

I=Vk\  (4) 

in  which  A;  is  a  length.  Similarly  the  second  moment  of  an  area 
A  may  be  expressed  in  the  form 

J=^2,  (5) 

and  likewise  for  any  magnitude  whose  second  moment  is  taken. 
The  length  k  as  defined  by  (4)  or  (5)  is  called  the  radius  of  gyra- 
tion of  the  given  magnitude  with  respect  to  the  reference  plane 
or  axis. 

Ex.  1.  Find  the  second  moment  and  the  radius  of  gyration  of  the  area  of 
a  semicircle  of  radius  a  with  respect  to  the  bounding  diameter. 

Taking  the  origin  at  the  center,  and  the  bounding  diameter  as  the  axis  of 
x,  we  have,  by  transforming  to  polar  coordinates,  y  =  p  sin  0,  A  A  =  p  Ad  Ap. 

Therefore,  Ix  =  \  "  y  p2 sin2  6  ■  pdddp  =  \  vat, 

and  *a  =  I*=l£^:=la», 

A     lira?     4 

whence  k  =  \  a. 

Ex.  2.  Find  the  second  moment  of  a  straight  rod  or  wire  of  length  I  about 
one  end,  assuming  that  the  density  varies  as  the  distance  from  that  end. 

Consider  the  rod  as  lying  in  the  X-axis  with  the  end  about  which  the  mo- 
ment is  taken  at  the  origin.  At  a  distance  x  from  the  origin  the  density  is 
kx  and  the  mass  of  an  element  of  length  Ax  is  kx  Ax.  The  second  moment 
of  this  mass  element  about  the  origin  is  x2  •  kx  Ax  ;  hence  the  second  moment 
of  the  rod  is 

7=    L    y.x2-kxAx=  C  kx*  dx  =  \  kl*. 


The  mass  of  the  rod  is  m  =  f  kx  dx  =  \  kl?. 

Therefore  &  =  -  =  \  P. 

m     2 


282  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 


EXERCISES 

Find  the  second  moment  and  the  square  of  the  radius  of  gyration  in  the 
following  cases. 

1.  A  rectangle  of  width  b  and  length  h  about  an  axis  coinciding  with  the 
short  side  b. 

2.  The  same  rectangle  about  an  axis  through  the  center  parallel  to  the 
side  b. 

3.  A  circle  of  radius  a  about  an  axis  through  the  center  perpendicular  to 
the  plane  of  the  circle. 

4.  The  same  circle  as  in  Ex.  3,  with  the  density  varying  as  the  distance 
from  the  center. 

5.  A  triangular  area  of  base  b  and  height  h  about  an  axis  through  the 
vertex  parallel  to  the  base. 

6.  A  semicircumference  of  a  circle  of  radius  a  about  the  diameter. 

7.  The  area  between  the  cycloid  x  =  a(d  —  sin  0),  y  =  a(l  —  cos  0)  and 
the  X-axis  about  the  X-axis. 

8.  The  cycloidal  curve  about  the  X-axis. 

9.  A  solid  sphere  of  radius  a  with  respect  to  a  diametral  plane. 

10.  The  same  sphere  as  in  Ex.  9,  assuming  that  the  density  varies  as  the 
distance  from  the  diametral  plane. 

139.  General  theorems  relating  to  second  moments.  The  follow- 
ing general  theorems  are  useful  in  the  determination  of  second 
moments  and  radii  of  gyration. 

Theorem  I.  If  a  solid  (or  homogeneous  mass)  is  divided  by  a 
plane  of  symmetry,  the  second  moment  of  the  entire  solid  (or  mass) 
with  respect  to  this  plane  is  double  that  of  the  part  on  one  side  of  the 
plane  of  symmetry. 

This  theorem  is  easily  established.  For  any  element  AF  on 
one  side  of  the  plane  at  a  distance  y  there  is  a  corresponding  equal 
element  on  the  other  side  at  a  distance  —  y.  The  second  moment 
of  one  element  is  y2  AV,  that  of  both  is  y2 AF+( -  y)2AV=  2  y2  AF. 
Since  all  the  elements  of  the  solid  can  thus  be  arranged  in  pairs, 
the  theorem  follows  at  once.  The  theorem  holds  also  for  the 
second  moment  of  a  plane  surface  with  respect  to  an  axis  of 
symmetry. 


Art.  139] 


SECOND   MOMENTS 


283 


Theorem  II.  The  second  moment  of  a  solid  {or  mass)  ivith 
respect  to  a  line  is  the  sum  of  the  second  moments  with  respect  to 
two  planes  which  intersect  at  right  angles  in  the  line. 

Consider  the  given  line  as  the  X-axis,  and  take  the  XZ-  and 
X  F-planes  as  the  intersecting  planes.  A  point  P  in  a  chosen 
volume  element  has  the  coordinates  (x,  y,  z).  Its  distance  from 
the  X-axis  is  V#2  +  z2,  and  by  the  definition  the  second  moment 
of  the  solid  with  respect  to  this  axis  is 


I, 


--f(y2  +  z2)dV. 


But 
hence, 


JV  dV=  Ixz,  and  J>  dV=  Ixy ; 
Ix  =  Ixz  +  Ixy. 


(1) 


A  similar  theorem  applies  to  plane  areas.  Referring  to  Fig. 
89,  it  is  clear  that  the  sum  of  Ix  and  Iy,  the  second  moments  of 
the  plane  area  with  respect  to  the  axes  of  x  and  y,  respectively,  is 
the  second  moment  of  the  area  with  respect  to  an  axis  through 
the  origin  0  perpendicular  to  the  plane  of  the  figure. 

.  Theorem  III.  The  second  moment  of  a  solid  (or  mass)  with 
respect  to  an  axis  is  equal  to  its  second  moment  with  respect  to  a 
parallel  axis  through  the  centroid 
plus  the  product  of  the  volume  (or 
mass)  of  the  solid  and  the  square  of 
the  distance  between  the  axes. 

Let  the  given  axis  pierce  the 
plane  of  the  page  at  0,  Fig.  93, 
and  suppose  Og  to  be  the  trace  of 
the  parallel  axis  through  the  cen- 
troid. Take  the  line  00g  as  the  FlG-  93- 
X-axis,  and  let  a  denote  the  distance  00g.  Referred  to  Og  as  an 
origin,  the  point  P  in  a  given  volume  element  has  the  coordinates 
(x,  y) ;  referred  to  0  as  origin,  the  coordinates  are  (x  +  a,  y). 
The  second  moment  of  the  solid  with  respect  to  the  axis  through 
Og  is  therefore 


Y 

Y-a— 

4 — *- 

o 


X 


Ig=f(x*  +  f)dV, 


284  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

while  the  second  moment  with  respect  to  the  axis  through  0  is 
I=fi(x  +  ay+y*]dY 

=  C(x2  +  y2)dV+a2CdV+2aCxdV 

=  Ig  +  d2V+2aCxdV. 

Now  J  x  d  V  is  the  first  moment  of  the  solid  about  the  axis  through 

Og\   hence,    since  this    axis    contains    the    centroid,   jxdV=0, 
and  we  have  finally 

I=Ig  +  a*V.  (2) 

Dividing  both  members  of  (2)  by  V,  we  get 

k2  =  kg2  +  a2.  (3) 

The  student  may  show  that  the  theorem  holds  for  parallel 
planes,  one  of  which  contains  the  centroid. 

Theorem  III  is  specially  useful  in  finding  the  second  moments 
of  volume,  as  illustrated  by  Ex.  2  given  below 

Theorem  IV.  If  kx,  k2,  •••,  kn  are  the  radii  of  gyration  of  n  vol-% 
umes  (or  areas,  or  masses)  with  respect  to  a  line  or  plane,  the  radius 
of  gyration  k  of  the  entire  system  with  respect  to  the  line  or  plane  is 
given  by 

k2  =  vjc£  +  vjc£±  •••  +vnK2_ 

This  theorem  is  useful  in  finding  the  second  moment  of  an  area 
or  volume  which  can  be  conveniently  divided  into  parts.  The 
proof  is  similar  to  that  of  Theorem  III,  Art.  137. 

Ex.  1.    Find  the  radius  of  gyration  of  a  circle  of  radius  a  about  a  tangent. 

From  Theorem  I,  the  radius  of  gyration  about  a  diameter  is  the  same  as 
that  for  a  semicircle  ;  hence  from  Ex.  1,  Art.  138,  we  have  kg2  =  \  a2.  Using 
Theorem  III,  we  obtain  therefore 

k2  =  \  a2  +  a2  =  |  a2. 

Ex.  2.  Find  the  second  moment  of  a  right  circular  cone  with  respect  to  a 
line  through  the  vertex  perpendicular  to  the  axis  of  the  solid,  Fig.  94. 

Take  the  axis  of  the  cone  as  the  Z-axis,  and  choose  the  vertex  as  the 


Art.  139] 


RADIUS   OF   GYRATION 


285 


origin.     Let  h  be  the  altitude  and  b  the  radius  of  the  base.     Consider  a  slice 
of  the  cone  parallel  to  the  XF-plane  at  a  distance  z  from  it.     The  radius  of 

the   slice  is  -z.  its  thickness  is  Az,  and  its 
h 

second  moment  with  respect  to  a  diametral 

axis  X',  parallel  to  OX,  is  approximately 


z*Az. 


According  to  Theorem  III,  the  second  mo- 
ment of  this  slice  with  respect  to  OX  is 
therefore 

1  r-  z*Az  +  *V-.sPAz-z* 


4      W 


h* 


Fig.  94. 


The  second  moment  of  the  cone  about  OX  is  the  limit  of  the  sum  of  terms  ot 
this  type  ;  that  is, 

5 


A*V4*»       J  Jo  Uh*        J 

The  radius  of  gyration  is  given  by  the  equation 

5   V4^2        /      20 v  J 


EXERCISES 
In  the  following  examples  determine  the  radii  of  gyration. 

1.  A  square  whose  side  is  a  :  (a)  about  an  axis  through  the  center  per- 
pendicular to  its  plane  ;  (6)  about  a  diagonal. 

Suggestion  :  Make  use  of  Theorem  II. 

2.  An  ellipse  with  semiaxes  a  and  b  :  (a)  about  the  major  axis  ;  (6)  about 
the  minor  axis ;  (c)  for  a  tangent  at  the  end  of  the  minor  axis  ;  (d)  for  a 
centroidal  axis  perpendicular  to  the  plane  of  the  ellipse. 

3.  A  ring  bounded  by  circles  of  radii  a\  and  a2,  respectively,  about  a 
diameter.    Use  Theorem  IV. 

4.  A  sphere  of  radius  a,  about  a  diameter. 

5.  A  right  circular  cone :  (a)  about  a  plane  containing  the  axis ; 
(&)  about  the  axis.    Take  a  as  base  radius  and  h  as  altitude. 

6.  A  right  circular  cylinder  :  (a)  about  its  axis  ;  (b)  about  a  generating 
element ;  (c)  about  a  diameter  of  one  base. 


286 


MULTIPLE   INTEGRALS.     APPLICATIONS    [Chap.  XIV. 


7.    Show  that  the  radius  of  gyration  of  a  right  circular  cylinder  about  its 

axis  is  the  same  as  that  of  the  circu- 
lar cross  section  about  the  same  axis. 
Deduce  a  similar  principle  for  pris- 
matic solids  in  general. 

8.  Find  the  position  of  the  axis  xc 
passing  through  the  centroid  of  the 
cross  section  shown  in  Fig.  95. 

9.  Determine  the  radius  of  gyra- 
tion of  the  cross  section,  Fig.  95  : 

(a)  with  respect  to  the  axis  OX; 
(&)  with  respect  to  the  axis  xc. 

140.  Illustrative  examples.  The  following  illustrative  prob- 
lems are  introduced  to  give  the  student  further  practice  in  setting 
up  the  definite  integrals  involved  in  various  summations. 

Ex.  1.  Find  the  total  pressure  on  a  circular  disk  of  radius  a  held  in  a 
vertical  position  below  the  surface  of  a  liquid. 

According  to  the  laws  of  hydrostatics,  the  intensity  of  liquid  pressure  is 
proportional  to  the  distance 
below  the  liquid  surface.  Tak- 
ing the  polar  element  of  area 
pAdAp,  Fig.  96,  the  pressure 
on  this  element  is  ky  p  Ad  A/>, 
where  &  is  a  constant ;  hence 
the  total  pressure  is  given  by  the 

double  integral  k\  j  y  p  dd  dp 

with  appropriate  limits  of  in- 
tegration. Evidently  the  limits 
for  p  are  0  and  a,  and  the  area 
of  the  semicircle  on  one  side 
of    the   vertical   diameter  will 


be  included  if 


=.  and  i 
2  2 


are 


Fig.  96. 


taken  as  the  limits  for  6.  It  should  be  noted  that  while  the  total  pressures 
on  the  semicircles  on  the  two  sides  of  the  vertical  diameter  are  equal,  the 
pressures  on  the  upper  and  lower  semicircles  are  not  equal.  Now  taking 
y  =  h  —  p  sin  0,  we  have  for  the  total  pressure 


P  =  2  k  f  w  (h  —  p  sin  6)p  dp  dd  -  ircPkh. 


Art.  140] 


ILLUSTRATIVE   EXAMPLES 


287 


The  mean  intensity  of  pressure  is  therefore  ircfikh  ■*■  ira2  =  kh,  which  is  the 
intensity  at  the  center  of  the  disk. 

Ex.  2.     Find  the  volume  of  liquid  that  will  flow  in  one  second  through  a 
rectangular  orifice  of  width  b,  Fig.  97. 

It  is  known  from  hydromechanics  that  the 
velocity  v  with  which  a  liquid  flows  through  a 
small  orifice  at  a  distance  h  below  the  liquid 
surface  is  given  by  the  equation  v  =  V2  gh ; 
therefore  the  volume  flowing  through  an  orifice 
having  the  area  A  A  is  AQ  =  V2~gh  A  A.  When 
the  orifice  is  large  the  height  h  is  different  at 
different  parts  of  the  orifice  and  to  determine  the  FlG"  97# 

total  volume  flowing  we  must  sum  the  volumes  flowing  through  the  elements 

into  which  the  area  of  the  orifice  is  divided.     Thus  Q  =  L^P  V'2gh  AA.     For 

the  rectangular  orifice  in  question  we  naturally  take  the  rectangular  ele- 
ment Ax  Ah ;  therefore 

•6 


Q 


Jh.   Jo 


2  gh  clx  dh 


f  bV^/thot 


fti*). 


Because  of  contraction  of  the  jet  and  friction,  the  discharge  determined 
experimentally  is  smaller  than  that  calculated. 


EXERCISES 

1.  Find  the  total  liquid  pressure  on  a  submerged  vertical  plate  of  width  6. 
(See  Fig.  97.)     Find  also  the  mean  intensity  of  the  pressure  on  the  plate. 

2.  Find  the  liquid  pressure  on  a  vertical  triangular  plate  having  a  vertex 
at  the  liquid  surface  and  the  opposite  side  horizontal. 

3.  In  the  preceding  examples  of  liquid  pressure,  show  that  the  total 
pressure  is  equal  to  the  area  of  the  plate  multiplied  by  the  intensity  of  pres- 
sure at  the  centroid  of  the  plate.  Prove  that  this  statement  holds  for  any 
submerged  plane  surface. 

4.  Write  the  definite  integral  with  proper  limits  of  integration  that  gives 
the  total  liquid  pressure  on  a  submerged  vertical  disk  with  an  elliptical  out- 
line, the  major  axis  being  horizontal. 

5.  Find  an  expression  for  the  volume  of  liquid  flowing  through  a  trian- 
gular notch,  that  is,  a  triangular  orifice  with  its  base  in  the  liquid  surface. 
Take  b  and  h  as  the  base  and  altitude  respectively. 

6.  Find  the  volume  of  liquid  flowing  through  a  semicircular  orifice,  radius 
a,  with  the  diameter  in  the  liquid  surface. 

7.  Set  up  the  definite  integral  that  gives  the  volume  flowing  through  a 
circular  orifice,  radius  a,  the  center  being  at  a  distance  h  below  the  liquid 
surface.     Take  (a)  rectangular  coordinates,  (6)  polar  coordinates. 


288  MULTIPLE   INTEGRALS.     APPLICATIONS     [Chap.  XIV. 

8.  Find  the  definite  integral  that  gives  the  second  moment  of  the  volume 

n/2        y'2        yi 

of  the  ellipsoid  —  +  —  H —  =1  with  respect  to  an  axis  parallel  to  the  Z-axis 
a2     b2     c2 

and  passing  through  the  point  (a,  0,  0) . 

Suggestion  :    Take  sections  parallel  to  the  FZ-plane  and  use  theorem  III, 

Art.  139. 

9.  A  cylindrical  vessel  is  filled  with  water  to  a  height  h.  A  small  orifice 
of  area  a  is  opened  in  the  lower  base,  the  water  flows  out,  and  since  the 
water  level  is  falling  continuously,  the  velocity  of  flow  varies  continuously. 
Denoting  by  A  the  area  of  the  cylindrical  cross  section,  find  an  expression  for 
the  time  required  to  empty  the  cylinder. 

10.   Find  an  expression  for  the  time  required  to  empty  a  hemispherical 
bowl,  radius  r,  through  an  orifice  of  area  a  in  the  bottom. 


MISCELLANEOUS    EXERCISES 

Find  the  following  areas : 

1.  Between  the  parabola  y2  =  8  x  and  the  circle  x2  +  y2  =  9. 

1  2 

2.  The  whole  area  of  the  curve  l-\    +  (  2  )    =  1. 


©'+(8 


3.  The  area  of  the  loop  of  the  curve  x3  +  y3  =  3  axy . 

4.  One  loop  of  the  curve  p  =  a  sin  2  6. 

q  a 

5.  The  part  of  the  curve  p  =  a  sin3  ■ —  below  the  X-axis. 

a 

6.  The  parabola  p  =  a  sec2  -  between  the  vertex  and  the  latus  rectum. 

7.  Fir\d  the  volume  bounded  by  the  cylinder  {x  —  a)2  +  (y  —  b)2  =  r2, 
the  surface  xy  =  cz,  and  the  plane  z  =  0. 

8.  Find  the  volume  included  between  the  surface  a2x2  +  b'2z2  =  2 (ax  -f-  bz) y 
and  the  planes  y  =  ±  m. 

9.  Find  the  volume  generated  by  revolving  the  curve  p  =  a(l  4-  costf) 
about  the  initial  line. 

10.  Find  the  area  of  that  part  of  the  surface  of  a  sphere  x2  +  y2  +  z2  =  2  az 
lying  within  the  paraboloid  z  =  mx2  +  ny2. 

11.  By  extending  the  method  of  Art.  108,  show  that  the  length  of  a  curve 
in  space  is  given  by  the  formula 


-£++{£>'+{£)'* 


12.    From  the  formula  of  Ex.  11,  find  the  length  of  the  helix 

z  =  a  arc  cos  - ,     z  =  a  arc  sin  # , 
b  b 

from  z  =  0  to  z  —  m. 


Art.  140] 


MISCELLANEOUS  EXERCISES 


289 


13.  Find  the  area  of  the  curved  surface  of  a  right  cone  whose  base  is  the 
astroid  x*  +  y1  =  aJ,  and  whose  altitude  is  c. 

Suggestion  :  Taking  the  origin  at  the  vertex  of  the  cone,  the  equation  of 

the  surface  is  xJ  +  y*  =  I  -  z  j    • 

14.  The  axis  of  a  right  circular  cylinder  passes  through  the  center  of  a 
sphere.  The  radius  of  the  sphere  is  a,  that  of  the  cylinder  is  6,  (6<a). 
Find  the  volume  of  the  part  of  the  sphere  external  to  the  cylinder. 

15.  Find  the  volume  included  between  the  surface  xy  =  cz  and  the  planes 
x  =  0,  x  =  6,  y  —  0,  y  =  &,  z  —  0,  z  =  a. 

16.  Find  the  mean  density  of  a  right  circular  cone  whose  density  varies 
as  the  distance  from  the  base. 

17.  Find  the  mean  density  and  mass  of  a  solid  hemisphere  of  radius  a, 
assuming  that  the  density  varies  as  the  distance  from  a  tangent  plane  parallel 
to  the  base. 

18.  Using  the  theorems  of  Pappus,  find  the  volume  and  surface  of  a  ring 
formed  by  revolving  a  circle  of  radius  a  about  an  axis  at  a  distance  b  from 
the  center  of  the  circle,  b>  a. 

19.  A  helical  screw  thread  whose  cross  section  is  an  equilateral  triangle 
is  cut  on  a  cylinder.  The  radius  of  the  cylinder  to  the  root  of  the  thread  is  r, 
and  the  height  of  the  thread  is  a.  Find  by  the  theorem  of  Pappus  the  volume 
of  one  turn  of  the  thread. 

20.  Apply  the  theorem  of  Pappus  to  find  the  surface  and  volume  of  a 
right  circular  cone. 

21.  Using  the  theorem  of  Pappus,  show  that  an  element  of  the  volume  of 
a  surface  of  revolution  is  2  wp2  sin  <p  Ap  A<p,  and  that  the  volume  is  therefore 
given  by  the  formula 


T 


V  =  2  ir  I   j  p2  sin  0  dp  dcp, 

with  proper  limits  of  inte- 
gration. 

22.  Using  the  theorem 
of  Pappus,  deduce  the  gen- 
eral form  for  the  polar 
volume  element,  viz. 

AF=/>2sin0ApA</>A0. 

23.  Find  the  radius  of 

gyration  of  the  hollow  rectangle,  Fig.  98,  about  the  axis  Xc,  which  passes 
through  the  center  of  the  figure. 

24.  In  Fig.  99,  the  thickness  of  the  plates  (a,  a)  and  channels  (6,  b)  is  f  inch 


■ 

'mam. 

u 

t 

j 

fjp 

\ 

■ 

m 

■f 

J«- 


Fig. 


Fig 


290 


MULTIPLE   INTEGRALS.     APPLICATIONS      [Chap.  XIV. 


throughout.     Find  the  length  e  in  order  that  the  radius  of  gyration  with  re- 
spect to  the  axis  Y Y  shall  be  the  same  as  that  with  respect  to  the  axis  XX. 

25.  A  pipe  elbow,  Fig.  100,  is  to 
be  provided  with  a  foot  a  to  sup- 
port it  in  the  position  shown.  The 
vertical  line  MM  through  the  foot 
should  therefore  pass  through  the 
centroid  of  the  elbow.  Find  the 
distance  c. 

26.  The  following  graphical 
method  is  used  for  finding  the  radius 
of  gyration  of  the  cross  section  of  a 
hollow  cylindrical  column.*  Lay 
off  to  a  scale  of  1  inch  equal  to  4 
(or  40)  the  inner  and  outer  radii  as 
the  legs  of  a  right  triangle;  then 
the  hypotenuse  measured  to  a  scale 

of  1  inch  equal  to  1  (or  10)  is  the  radius  of  gyration  sought.     Give  a  proof  of 
this  construction. 

27.  By  particular   examples   verify  the    following  rule,  due  to  Routh  : 
For  homogeneous  masses  with  axes  of  symmetry,  the  square  of  the  radius 

of  gyration  is  },  J,  or  I  of  the  sum  of  the  squares  of  the  perpendicular  semi- 
axes,  according  as  the  mass  is  rectangular,  elliptic,  or  ellipsoidal. 

28.  From  the  second  moment  of  a  sphere  about  a  diameter  deduce  by 
differentiation  the  second  moment  of  a  spherical  shell  about  a  diameter. 

29.  Find  the  centroid  of  the  volume  OABCD,  Fig.  87. 

30.  (a)  For  a  surface  of  revolution  show  that  the  radius  of  gyration 
about  the  axis  of  revolution  is  given  by  the  equation 


Fig.  100. 


*2 


j  y*  ds 

\  yds 


where  the  X-axis  is  taken  as  the  axis  of  revolution,     (b)  Show  that  for  the 
corresponding  solid  of  revolution 


k- 


!■' 


Adx 


y 


*dx 


*  B.  F.  LaRue  in  Engineering  News,  Feb.  2,  1893. 


CHAPTER   XV 
INFINITE   SERIES 

141.  Fundamental  definitions.  From  the  study  of  algebra,  the 
student  is  already  familiar  with  the  elementary  properties  of  in- 
finite series.  In  the  present  chapter,  we  shall  recall  briefly  some 
of  the  more  important  of  those  properties  and  develop  still 
others. 

Let  u1}  t%,  uSf  •••  be  an  infinite  succession  of  values  following 
one  another  according  to  some  fixed  law.  We  denote  the  sum  of 
the  first  n  of  these  values  by  Sn,  that  is, 

#»  =  «i.+ *»  +  •••  +un-  (1) 

When  n  becomes  infinite,  we  have  the  infinite  series 

w14-t«2+«3+  ••••  (2) 

If  Sn  has  a  limit  as  n  becomes  infinite,  that  limit  is  called  the  sum 
of  the  infinite  series.     The  series  (2)  is  said  to  be  convergent  if  we 

have  L     Sn  =  A,  (3) 

n  =  oo 

where  A  is  a  definite  number ;  in  all  other  cases  the  series  is  said 
to  be  divergent.  As  n  increases  indefinitely,  S„  may  fail  to  have 
a  limit,  and  therefore  the  series  may  be  divergent,  either  because  Sn 
oscillates  between  two  numbers,  as  in  the  series 

1-1  +  1-1  +  1-1+  •», 

or  because  Sn  ultimately  exceeds  every  finite  number. 

The  terms  of  the  series  may  be  functions  of  some  variable  x. 
Then  the  series  is  said  to  converge  for  any  particular  value  of  this 
variable,  say  x  =  x0,  when,  if  x  is  replaced  by  x0  in  each  term,  we 

have  L    Sn(x0)=A. 

n  =  oo 

291 


292  INFINITE    SERIES  [Chap.  XV. 

If  the  corresponding  limits  exist  for  all  values  of  xina  certain 
interval  (a,  /8),  the  series  is  said  to  converge  throughout  the  inter- 
val and  to  define  a  function  in  the  interval.     We  may  then  write, 

f(x)  =  L    SJx),  a£x<fi. 

n  =  00 

Ex.  1.   Let  the  given  series  be 

1  +  x  +  x2  +  x8  +  ...  +  xn  +  .... 
We  have  then 

l  —  xn 


L  Sn(x)  =  L 


1-x 


This  limit  is  for  all  values  of  x  within  the  interval  —  1  <£<  +  1. 

1  —  x 

Hence  within  this  interval  the  series  defines  the  function 

It  is  to  be  noted  that  in  this  case  f(x)  is  defined  by  the  series  only  for  the 
interval  (—1,  1),  the  end  points  being  excluded.  For  |x|>l  the  series 
becomes  divergent  and  does  not  define  a  function. 

Ex.  2.    Consider  the  series 

1+|  +  |+...+2+.„. 


This  series  may  be  written  in  the  form 

i+i+fi+iV+fi+i+MU 

2      \3     4/      \6     6     7     8/ 

W      w  +  1  2(w-  1)/ 


The  sum  of  the  terms  in  each  of  these  parentheses  is  greater  than  J,  and  as  the 
number  m  of  such  groups  that  can  be  formed  from  the  given  series  is  indefi- 
nitely large,  we  have  T    a  .       T  ,        ,  N 

'       6  '  L  8n  >    L  (m  •  ^)  =  go  . 

n  =  oo  m  —  cc 

This  result  leads  to  an  important  observation ;  namely,  that  a  series  is  not 
necessarily  convergent  because  the  terms  themselves  decrease  and  approach 
zero  as  n  increases. 

If  a  series  containing  negative  terms  is  still  convergent  when 
all  of  the  negative  terms  are  taken  positively,  that  is,  when  only 
the  absolute  values  of  the  terms  are  considered,  the  series  is  said 
to  converge  absolutely  or  unconditionally.  If  the  series  is  con- 
vergent when  the  negative  terms  are  taken  with  their  proper  signs, 


Art.  142]  TESTS   OF  CONVERGENCE  293 

but  is  not  convergent  when  those  signs  are  taken  positively,  the 
series  is  said  to  converge  conditionally.     The  series 

2^3     4^5 

is  such  a  series.     On  the  other  hand,  the  series 

2     22     23 
is  an  absolutely  convergent  series,  since  it  converges  when  all 
terms  are  given  the  positive  sign. 

142.  Tests  of  convergence.  To  test  the  convergence  of  a  series, 
certain  criteria  are  necessary.  It  is  to  be  remembered  that 
whether  the  terms  of  the  series  are  constants  or  functions  of  a 
variable,  we  are  concerned  only  with  the  limit  of  the  sum  of  a 
finite  number  of  constant  terms  as  that  number  increases  indefi- 
nitely. For,  in  case  the  terms  are  functions  of  a  variable,  we 
either  substitute  a  constant  value  for  the  variable  and  then  pass 
to  the  limit,  or,  what  amounts  to  the  same  thing,  we  consider  for 
what  values  of  the  variable  the  series  has  a  limit.  It  is  impor- 
tant, however,  to  distinguish  between  convergence  at  a  point,  and 
convergence  throughout  an  interval. 

The  following  tests,  already  considered  in  algebra,  and  conse- 
quently needing  no  proof  here,  will  be  found  sufficient  for  series 
which  arise  in  ordinary  practice.* 

Theorem  I.     Comparison  test.     Given  a  series  of  positive  terms 

^1  +  ^2  +  ^3+    ••'    +**«+    •••• 

If  from  some  point  on  the  terms  of  this  series  are  never  greater  than 
the  corresponding  terms  of  a  known  convergent  series 

Vi  +  v2 -f  i's  +  •••  +vn+  ••• 
of  positive  terms,  then  the  given  series  is  convergent.     If  the  terms 
of  the  given  series  are  from  some  point  on  never  less  than  the  corre- 
sponding terms  of  a  known  divergent  series  of  positive  terms 

then  the  given  series  is  divergent. 


*  Compare  Rietz  and  Crathorne's  College  Algebra,  Chapter  XVI. 


294  INFINITE   SERIES  [Chap.  XV. 

.  Ex.  1.  Test  the  converge ncy  of  the  series 

From  Ex.  1  of  the  previous  article  we  have  the  convergent  series  (  |  x  |  <  1) 
l+z2  +  z3  +  z4+  -.  +  xn  +  -..  (2) 

For  x  =  |,  this  series  becomes 

Since  for  n  >  2,  2"-1  <  n  !,  and  each  term  of  (1)  after  the  second  is  less  than 
the  corresponding  term  of  (3);  therefore  the  series  (1)  must  converge. 

Ex.  2.     Given  the  series 

1+i+S+f+-.+sfei+"--  (4) 

Compare  the  series  with  the  series  whose  first  n  terms  are 

1+1+1+ ...  +J-=i(i +!+'!+ ...+iy  (5) 

2      4      6  2»      2V        2      3  n)  KJ 

By  Ex.  2  of  the  previous  article,  the  series  (5)  is  divergent.  Moreover,  since 
2  n  —  1  <  2  w,  each  term  of  the  given  series  (4)  is  greater  than  the  corre- 
sponding term  of  the  series  defined  by  (5)  ;  consequently,  the  series  (4)  is 
divergent. 

Theorem   II.     Ratio  test.     A  given  series 

ux-\-u2+  ...  -\-un-1  +  un  +  ••• 

u 
is  convergent  or  divergent,  according  as  the  limit      L     — —is  less  or 

greater  than  1.  n  =  °°  Wn~x 

Ex.  3.     Given  the  series 

1  +  x  +  ■  —  +  —  +  —  +     a;M~1      +  — +— .  (6) 

2!^3!^       T(*-l)l      n!  W 

We  have  then 

SB* 


X     -=2-=:      x    ^r- 


UJi-  —       r  Wl  T      * 


(11-1)1 

for  any  finite  value  of  x.     Hence  the  series  converges  for  all  finite  values  of  x. 

It  is  to  be  noted  that  this  second  criterion  applies  equally  well 
when  some  of  the  terms  of  the  series  are  negative.     It  gives  no 

test,  however,  when  the  limit  of  the  ratio  — *-  is  1 ;  in  such  cases 
other  methods  must  be  employed.  n_1 


Abt.  143]  POWER   SERIES  295 

Since  each  term  of  a  series  is  finite,  the  sum  of  any  finite  num- 
ber of  terms  is  also  finite,  and  consequently  the  convergency  or 
divergency  of  a  series  is  not  affected  by  omitting  a  finite  number 
of  terms. 

If  the  terms  of  the  given  series  are  alternately  positive  and 
negative,  the  following  is  a  convenient  test  of  convergence. 

Theorem  III.  An  infinite  series  in  which  the  terms  are  alternately 
positive  and  negative  is  convergent  if  its  terms  decrease  numerically 
and  approach  zero  as  a  limit  ivhen  n  increases  indefinitely.* 

143.  Power  series.  By  a  power  series  we  understand  a  series 
of  the  form     ^  +  v+^  +  i^+  ...  +ajf>+  •",  (1) 

where  a0,  al}  •••,  an,  etc.,  are  constants. 

Such  series  are  of  importance  in  mathematics  because  of  the 
frequency  with  which  they  occur  and  because  of  the  special 
properties  which  they  possess.  For  example,  it  is  not  always 
possible  to  obtain  the  integral  or  the  derivative  of  a  function  by 
integrating  or  differentiating  term  by  term  the  series  which  de- 
fines the  function.  A  power  series,  however,  has  this  important 
property  when  the  variable  is  restricted  to  a  proper  interval  of 
convergence.  Consequently,  if  f(x)  is  defined  by  the  power 
series  f^  =  ^  +  ^  +  ^  +  ...  +  ^  +  . .  m$ 

we  may  then  write 

|     f(x)dx=  J  *la0dx  +  CXlajxdx+  •••  +  j    ianx*dx  +  ••• 

and 

df(x)  _d(a0)     d(alx)  d(anx) 

dx  dx  dx  dx 

when  we  place  a  suitable  restriction  upon  the  value  of  x.f 

A  valuable  property  involving  the  convergence  of  a  power 
series  is  given  by  the  following  theorem. 


*  For  proof  of  this  theorem,  see  Rietz  and  Crathorne's  College  Algebra, 
p.  188. 

t  For  a  fuller  discussion  of  the  conditions  for  term-by-term  differentiation 
and  integration  of  a  series,  see  First  Course,  Chap.  XV. 


296  INFINITE   SERIES  [Chap.  XV. 

Theorem  T.     If  a  power  series  converges  for  x  =  a,  it  is  abso- 
lutely convergent  for  all  values  of  x  such  that  |  x  |<  |  a  |. 

Let  the  given  series  be  written  in  the  form 

«o  +  aJ-)  +  a2a2(*Y  +  ■  •  •  +  o^WsY  +  •  •  • ;  (2) 


then  the  series  of  coefficients 

a0,  «!«,  a2a2,  •••,  anan,  ••• 

must  decrease  indefinitely  in  absolute  value  since  the  given  series 
converges  for  x  =  a.  Let  M  be  equal  to  or  greater  than  the  abso- 
lute value  of  any  number  in  this  sequence.  Then  the  absolute 
values  of  the  terms  of  (2)  are  less  than  the  corresponding  terms 
of  the  geometric  series 

\        a      er      or  an 

which  converges  for  |aj|<|«|.  Hence  the  given  series  (1)  con- 
verges absolutely  for  |  x  \  <  |  a  | . 

144.  Maclaurin's  expansion  of  a  function  in  a  power  series.  It 
is  often  convenient  to  express  a  function  in  terms  of  a  series. 
A  power  series  is  very  serviceable  for  this  purpose,  because,  as 
we  have  seen,  such  series  may  be  differentiated  and  integrated 
term  by  term,  thus  obtaining  the  same  result  as  if  the  opera- 
tion had  been  performed  upon  the  function  itself.  In  the 
following  article  we  shall  discuss  two  methods  of  expanding 
functions  in  terms  of  a  power  series  by  making  use  of  the  prin- 
ciples of  calculus. 

Suppose  we  have  given  a  function  f(x)  which,  together  with  its 
derivatives,  is  continuous  in  the  vicinity  of  the  value  x  —  0.  If 
such  a  function  can  be  represented  by  a  power  series,  that  series 
must  be  of  the  form 

f(x)  =  A»  +  Alx  +  A2a?+  -.  +Anxn+  ...,  (1) 

where  the  coefficients  A0,  Ax,  A2,  •••,  An,  •••  are  to  be  determined. 
Since  this  is  a  power  series,  we  may  find  the  successive  deriva- 
tives of  the  function  f(x)  by  term-by-term  differentiation  of  the 
series.     The  following  identities  are  thus  derived : 


Art.  144]  EXPANSION   OF   A   FUNCTION  297 

f\x)  =  Ax  +  2  A2x  +  3  Atf  +  4  Atf  +  ... 
f"(x)  =2  A2  +  3  •  2  A3x  +  ±  •  3  A4x>  +  ... 
f"'(x)  =  3.2As  +  <l-3.2A4x+  ... 

r*(x) =4.3.2^+  - 


Substituting  #  =  0  in  (1)  and  in  each  of  the  above  identities,  we  have 

f(0)  =  A0,  f\0)  =  Al,  f"(0)  =  2\A2)  /'"(0)  =  3!^3,  .... 
Hence  the  successive  coefficients  are 

A=/(0),  A=m,  j,=qH,  A  =  «,  -.       (2) 

We  have  thus  the  values  of  the  unknown   coefficients   in   the 

assumed  expansion  in  terms  of  the  successive  derivatives  of  the 

given  function.     Substituting  these  values  in  that  expansion,  we 

have 

/(^)=/(0)4-/'(0)^  +  ^^^2  +  ^^^  +  ... 
-  .  •> . 

+rmxn+...,         (3) 

where  /"(0)  denotes  the  result  obtained  by  differentiating  the 
function  fix)  in  succession  n  times  and  substituting  x  =  0  in  the 
result.     The  above  series  is  called  Maclaurin's  series. 

For  any  given  function,  there  still  remains  to  be  determined  the 
interval  within  which  the  expansion  obtained  by  (3)  really  repre- 
sents that  function.  This  question  will  be  discussed  in  a  subsequent 
article.  Assuming  that  such  an  expansion  is  possible,  the  fol- 
lowing examples  will  illustrate  the  method  of  computing  the 
successive  coefficients  in  the  expansion.  When  the  successive 
derivatives  all  become  zero  from  some  point  on,  the  expansion 
has  a  finite  number  of  terms. 

Ex.  1.   Expand  f(x)  =  (1  +  x)m. 

We  have  f(x)  =  (1  +  x)m, 

/'(«)=  m(l  +*)«-*, 

f"(x)  =  m{m  -  1)  (1  +  x)*-2, 


f»(x)=m(m-  1)  ...(»—  »-M)(l  +x)m~n, 
whence  /(0)  =  1,  /'(0)  =  m,  /"(0)  =  m(m  -  1),  etc. 


298  INFINITE   SERIES  [Chap.  XV. 

Substituting  these  values  in  (1),  we  have,  when  m  is  negative  or  fractional, 
the  infinite  series 

(1  +  *)«  =  1  +  mx  +  m(m  ~Vx*  +  m(m  ~  *> <>  ~  2>  x*  +  ••  • 

2  !  3  ! 

t    W(W-1)-.(W-71+1)XM         _ 

w  !  ' 

which  converges  for  |  x  j  <  1. 

If  m  is  a  positive  integer,  the  series  terminates  with  first  m  -f  1  terms, 
since  all  of  the  higher  derivatives  vanish. 

Ex.  2.   Expand  sin  as  in  a  power  series. 

/(as)  =  sin  x,  /(0)  =  sin  0  =  0, 

f'(x)=cosx,  /'(0)=1, 

/"(«)  =  -  sin  as,  /"(0)  =  0, 

f't'(x)= -C08S,  /'"(0)  =  -l, 

/IV(z)  =  sin  sb,  f™(0)  =  0, 

etc.  etc. 

Hence,  sin*  =  0  +  x  +  0  —  — +  0  +  — +  0  — — +  — 

3 !  5!  7 ! 

-7*3  'Y'h  npl 

=  x—  —  +  —  —  —  H — . 
3!^5!      7!^ 

The  interval  of  convergence  for  this  series  is  (—  oo,  +  qo). 


EXERCISES 

Expand  in  power  series  the  following  functions,  assuming  that  such  ex- 
pansions are  possible. 

1.  ex.                          2.   cos  a;.                      3.  ax.                         4.    esinx. 

5.  log  (1  +  ex).                       6.  log(l+x).  7.  arc  sin  a. 

8.  arc  tan  x.                            9.  (cos  0)n.  10.  e9  sec  6. 

11.  sec  6  (to  four  terms).      12.  «*»•»*•.  13.  log  (a;  +  Vl  +ic2). 

14.  log  cos  6.                          15.  l(ex  +  e~x).  16.  log(l  —  x  +  x2). 

145.  Taylor's  expansion.  In  the  last  article,  we  studied  the 
expansion  of  a  given  function  f(x)  in  the  vicinity  of  the  value 
x  =  0.  The  method  may  be  easily  extended  to  the  expansion  in 
the  neighborhood  of  any  point  x  =  a,  provided  the  given  function 
and  its  successive  derivatives  are  continuous.  All  that  is  neces- 
sary is  to  assume  the  expansion  of  the  form 

f(x)=A{)  +  Al(x-a)  +  A2(x-cif  +  ...  +An(x-a)n+  •••     (1) 


Art.  145]  TAYLOR'S  EXPANSION  299 

and  proceed  precisely  as  in  Art.  144.  The  resulting  form  of  the 
expansion  is 

+/jj^»-a)»  +  .«1  (2) 

which  holds  for  all  values  of  the  variable  within  the  interval  of 
equivalence.  The  series  resulting  from  this  expansion  is  known 
as  Taylor's  series. 

If  in  (2)  we  replace  x  by  (x-\-  a),  we  have 

f(x  +  a)=f(a)+f\a)x  +  £Mx?+...+£Mx«+...!      (3) 

Z\  n 

which  is  a  form  in  which  Taylor's  series  is  frequently  written. 
If  x  and  a  are  interchanged,  the  expansion  takes  the  form 

f(a  +  x)=f(x)+fXx)a+£^a*+  ...  +  ££)«•  +  ....         (4) 

Forms  (3)  and  (4)  are  useful  when  it  is  desired  to  expand  a 
function  of  the  sum  of  two  numbers  in  powers  of  one  of  them. 

Ex.  1.   Expand  ex  in  powers  of  x  —  1. 

We  have                   f(x)  =  e*,  /(l)  =  e, 

/'(«)=«*,  /'(1)=  0, 

/"(*)=*>,  /"(!)  =  «, 

etc.  etc. 

Hence,  e*  =  e[~l  +(«-  1)  +  ^=^  +  C^zlII3  +  ...]. 

Ex.  2.   Express  3  ar3  —  5  x2  +  8  x  —  5  in  powers  of  x  —  2. 
In  this  case 

/(*)  =  3  Xs  -  5  a-2  +  8  x  -  5,  /(2)  =  15, 

/'(as)  =  9z2-10  x  +8,  /'(2)  =  24, 

/"(*)  =  18*  -10,  /"(2)  =  2G, 

/'"(«)  =  18,  /'"(2)=18, 

•    /IV(*)=0,  /IV(2)=0. 

Hence,  we  have 

/(*)  =  /(2)+/'(2)(z  -  2)  +  ^^(x  -  2)2  +  £^p-(x  -  2)8, 

or  3  x^  -  5  *2  +  8  x  -  5  =  15  +  24(x  -  2)  +  13(z  -  2)2  +  3 (a;  -  2)3. 


300  INFINITE   SERIES  [Chap.  XV. 

Ex.  3.     Develop  log  (a;  4-  h)  in  powers  of  h. 

We  have  /(x  -f  h)  =  log  (x  +  A),  /(*)  =  logx, 

/w-i.  /"(*)=-£,    . 


Substituting  in  (3),  we  obtain 

log  (x  +  A)  =  log  x  +  *  -  ^  +  -^  -  -^  +  ».. 
x      2x2      3x3      4x4 
For  x  =  1,  we  have 

log(l  +  A)  =  A-f  +  f-f+.... 

Ex.  4.     Expand  sin  (x  +  y)  and  derive  the  formula 

sin  (x  +  y)  =  sin  x  cos  ?/  +  cos  x  sin  y. 

We  have  /(x)  =  sin  x, 

/'  (as)  =  cos  x, 

/"(*)  =-sinx, 

f'"(x)  =  —  cos  x,  etc. 

Substituting  in  (4),  the  result  is 

.     ,  .        .  V2    .  V8  V4    .  V5 

sin  (x  -f  y)  =  sin  x  +  y  cos  x  —  |-y  sin  x  —  ^-j  cos  x  +  j7  sin  x  +  ~  cos  x  —  ••  • 

\        2!  ^4!      6!^       ) 

+co§«('-fr+gi-n+"") 

=  sin  x  cos  y  +  cos  x  sin  y. 

Ex.  5.     If  /(x)  -  5  x8  -  4  x2  +  18  x  -  7,  find/(x  -  2)  by  Taylor's  expan- 
sion. 

We  have  here  /'(x)  =  15  x2  -  8  x  +  18, 

/"(x)  =  30x-8, 
/'"(*)  =80, 
Hence  from  (1),  /•*(*)  =  0. 

f(x  -  2)  =  /(*)  +  (-  2)  /'(a)  +  ^|I2 /"(x)  +  ^-8  /'»(«) 
=  6x8-'4x2-fl8x-8 
-  30  x2  +  16  x  -  36 
+  60  x  -  16 
-40 
=  5  x3  -  34  x2  +  94  x  -  100. 


Arts.  145,  146]  TAYLOR'S  THEOREM  301 

EXERCISES 

Develop  the  following  functions. 
1.   ex+h.  2.    (x  +  y)m.  3.    (x  +  y)«. 

4.   Arc  sin  (x  +  h)  to  four  terms.  5.    Log  sin  (x  +  h). 

6.  Find  /(a  +  S),  when  f(x)  =  x3  -  4  x  +  7. 

7.  Find  /(x  -  1),  when  /(x)  =  x2  +  7  x  -  5. 

8.  Show  that  log  (x  +  VI  +  x'2)  =  x  -  —  +  -^ .... 

oV  '  2-3      2-4.5 

9.  Expand  sin  x  in  powers  of  x  —  a. 

10.  Express  5  x3  —  6  x2  +  x  —  10  in  powers  of  x  —  1  ;  also  in  powers  of 
x  —  3.  Verify  the  results. 

11.  Expand  logx  in  powers  of  x  —  1.     Find  the  interval  of  convergence. 

12.  Expand  -  in  powers  of  x  —  a  and  determine  the  interval  of  convergence. 

146.  Taylor's  theorem.  Maclaurin's  theorem.  In  discussing  the 
expansion  of  a  function  in  terms  of  a  power  series,  we  have  as- 
sumed that  the  series  obtained  actually  represents  the  function. 
We  shall  now  see  under  what  conditions  this  is  true.     We  may 

write  fix)  =  Sn(x)  +  Rn(x),  (1) 

where /(x)  is  the  given  function.     If  the  infinite  series  given  by 

L    Sn  (x)  represents  f(x)  in  a  given  interval,  then  for  all  values 

n  =00 

of  x  in  that  interval,  we  must  have 

L    Bn(x)=0;  (2) 

n  =  oo 

for,  Rn(®)  represents  the  difference  between  the  given  function 
and  the  sum  of  the  first  n  terms  of  the  series,  that  is,  the  error 
involved  by  stopping  the  expansion  with  n  terms.  It  is  conven- 
ient to  have  the  value  of  Rn  (x)  expressed  in  terms  of  the  deriva- 
tives of  f(x).  This  value  can  be  obtained  for  Taylor's  expansion 
as  follows. 

From  equation  (1),  we  have 

f(x)=f(a)+f(a)(x-a)+^(x-ay+  ... 


802  INFINITE   SERIES  [Chap.  XV. 

Since  Bn(x,  a)  contains  the  factor  * — *-,  we  may  write  it  in 

ft! 

the  form  \x~a)  ^^  a\     Replacing  Rn(x,  a)  by  this  expression 
n ! 

and  transposing,  we  have 
f(x)-f(a)-f(a)(x-a)-^(x-af gT^l  (*_„)»-> 

-jj&-2l(x~ay  =  0.  (4) 

To  find  the  value  of  <f>(x,  a)  in  terms  of  the  derivatives  of  f(x), 
we  shall  consider  the  function 

F(z)=f(x)-f(z)-f'(z)(x-z)—£^(x-zy--  - 

f—n(*) ,         N„_i      <t>(x,d),         ..  ,~. 

-fc^ijp     "J        ^Ti    {x~z)  () 

F(z)  satisfies  the  conditions  of  Rolle's  theorem  in  that  it  pos- 
sesses a  derivative  for  each  value  of  z,  where  a  ^  z  <J  x,  and  van- 
ishes for  z  —  x  and  z  =  a.  Differentiating  (5)  with  respect  to  z,  we 
have 

*"(*)-  -f'(z)+f(z)-f"(z)(x-z)+f"(z)(x-z)-  ... 

The  terms  in  the  second  member  of  this  equation  combine  in 
pairs  and  the  final  result  is 

Since  F(z)  satisfies  Rolle's  theorem,  F'(z)  vanishes  for  some  value 
of  z  between  a  and  x,  say  xv     We  have  then  from  (7) 

4>(x,a)=r(*i),  (8) 

and 

BL  »-£&&(•'•- o)».  (9) 


Art.  147]  INTEGRATION   OF  SERIES  303 

Replacing  Ra(x)  Dv  ^s  value  as  given  in  (9),  we  may  now  write 
(3)  in  the  following  form 

/(^)  =  /(a)+/'(a)(a?-tt)  +  ^^ 

+  ^T  (*-«)"•  (10) 

This  formula  is  known  as  Taylor's  theorem. 

We  are  now  in  a  position  to  determine  the  interval  within 
which  the  expansion  represents  the  given  function  by  determin- 
ing the  range  of  values  of  x  for  which  (2)  holds.  In  the  simple 
cases  which  will  come  under  consideration  this  interval  will 
usually  coincide  with  the  interval  of  convergence  of  the  series. 

We  have  seen  that  Maclaurin's  expansion  is  a  special  case 
of  Taylor's.  By  putting  a  =  0,  we  have  as  the  value  of  Bn(x)  in 
Maclaurin's  series 

Mn(x)  =£{&&,     0<x,<x.  (11) 

Consequently  f(x)  is  given  by  the  relation 

/(*)  =  /(0)  +  /W*+^**+  -  +^M,»-.+^,. 

which  is  known  as  Maclaurin's  theorem. 

By  means  of  this  theorem  we  may  determine  the  interval  within 
which  Maclaurin's  expansion  represents  the  function.  In  any 
given  case  we  have  only  to  determine  the  range  of  values  of  x  for 

which  Bn=    jSu  xn  has  the  limit  zero  as  n  becomes  infinite. 

147.  Integration  and  differentiation  of  series.  It  is  sometimes 
possible  to  expand  a  function  into  an  infinite  series  by  means 
of  term-by-term  integration  or  differentiation  of  a  known  series. 

Again  if  a  given  integral    J  f(x)  dx  cannot  be  evaluated  by  the 

ordinary  exact  methods  of  integration,  it  may  be  possible  to  de- 
velop f(x)  into  an  infinite  series  and  integrate  term  by  term.  By 
taking  a  sufficient  number  of  terms  of  the  series  resulting  from 
the  integration,  we  may  approximate  the  given  integral  to  any 
desired  degree  of  accuracy.  Of  course  the  series  so  treated  must 
be  such  as  can  be  differentiated  or  integrated  term  by  term. 


304  INFINITE   SERIES  [Chap.  XV. 

Even  when  the  function  can  be  integrated  directly  it  is  some- 
times convenient  to  use  the  method  just  described,  for  the  series 
may  be  more  easily  handled  in  the  subsequent  operations  than  a 
complicated  integral. 

Ex.  1.    For  —  1  <  x  <  1,  we  have 

1 


l  +  x 


=  l-x  +  x2-x?  + 


Hence,         f  *_^_  =  C*dx-  Cxdx  +  Cx2dx-  Cx^dx  +  .... 
Jo  1  +  x     Jo  Jo  Jo  Jo 


r2         r3         r4 

or  l0g(l+z)  =  x-|+|-|  + 


Ex.  2.     For-l<sc<l, 

_!—  =  i  +  x  +  x2  +  xs+  •-.. 

1  —  X 

Differentiating  both  members,  we  obtain 
1 


-  =  1  +2z  +  3se2  +  4z3+  .... 

(1_Z)2 

Likewise,  a  second  differentiation  gives 

_J_?=i(1.2  +  2.S*  +  8.4*»+...), 

and  in  general  we  have 

Tj-^  =  (1  -  *)~m  =  Umx  +  ^±HX2  +  m(m  +  l)(m  +  2)  g8  +  _. 

Ex.  3.     The  perimeter  of  an  ellipse,  which  has  a  for  semimajor  axis  and 
e  for  eccentricity,  is  given  by  the  integral 


4a  f    Vl-  e2 sin2 <pd<f>. 


This  integral  can  be  evaluated  approximately  by  expansion  in  series.    Thus 
we  have 

Vl  -e2Sin2<6=  l_le2sm<J0_I   .    I  e4sm40_!    .    *    .    ?  e6sin60_    .. 

2  24  ^246 

which  is  a  power  series.     Integrating  term  by  term,  we  have 

7T  TV  7T  IT 

f 2  Vl  -e2  sin2  <f>d<f>  =  (*  d<p--e2  ( *  sin2 <p d<p  -  -  .  -  e4  f  sin^ety-  •••. 

Evaluating  these  integrals  separately,  we  get  for  the  required  perimeter  the 
expression 

L        \2)    1      \2     ij   S      V*     4     «/    6  J 


Art.  147]  INTEGRATION  OF   SERIES  305 

EXERCISES 

1.  From  the  known  series 

1  -  x2  +  x4  —  x6  +  ••-, 

which  defines  the  function for  —  1  <  x  <  1,  derive  the  series  for  arc  tan  x. 

1  +  x2 

2.  For  -  l<x<  1, 


VIT^  2  2-4  2-4-6 

Derive  from  this  relation  a  series  for  arc  sin  x. 

3.  Derive  the  series  for  cos  x  by  differentiating  that  for  sin  x. 

4.  Showthat  f11°g(1+^=l-l  +  l-l  +  .... 

Jo  x  l2      22      32      42 

_     „  f1  since  da;  .  -  .. 

5.  Express   \    as  an  infinite  series. 

6.  Express  I as  an  infinite  series. 

7.  Find  (*"'       dx       to  five  figures. 
J9    >/i  _  ^2 


8.   Evaluate 


VI  -x2 


1      <?x 


x 


9.    The  time  of  oscillation  of  a  pendulum  of  length  L  is  given  by  the 
expression  n 

T^J^C  —  ** 

y  9  Jo    Vl  -  k2  sin2  <p 

Integrate  in  series,  and  derive  an  approximate  expression  for  T  when  k  is 
small. 

10.-  Expand  f — ^—  in  series. 
*  Vsin  x 

11.  Evaluate   \  e~x2  dx  by  expansion  in  series.     This  integral  is  of  funda- 
mental importance  in  the  theory  of  probability. 

12.  Show  that 

C *± =    *•    Tl     1k2      8  Jfc*      5  y--1 

J°    («2cos2*  +  62sin2^      2<^L       4  64  256  J' 

where  A;2  =  1  -  -0  • 
a2 


306  INFINITE   SERIES  [Chap.  XV. 

148.  Use  of  series  in  computation.  Infinite  series  may  be  used 
advantageously  in  the  computation  of  certain  constants,  loga- 
rithms, trigonometric  functions,  and  roots  of  numbers ;  also  in  the 
derivation  of  certain  useful  approximations. 

I.  Computation  of  e. 

We  have  e*  =  l  +  »  +  ^  +|^+  •••-  (1) 

Therefore  for  x  =  1, 

e  =  l  +  l+  —  +  —  +  —  +  — ,  (2) 

2!3!4!  J 

whence,  taking  a  sufficient  number  of  terms,  e  =  2.7182818- ••. 

II.  Computation  ofir. 
From  the  expansion 

23      2.45      2-4.6  7  '  W 

which  holds  for  —  1  <  x  <  1,  we  get  for  x  =  7 

-  =  -  =  -4--     -     f-Y  +  t     ?     1     /lY-*-  (ft 

2     6      2~t~2*3*\2y       2*4*5     \2J         "'       K) 

whence  ir  =  3.14159-  ••. 

III.  Extraction  of  roots. 

i 

-  /  h  \  n  1 

We  have        (an  ±b)n  =  a(l  ± -)   =a(l±x)n,  (5) 

where  #  =  — .     Developing  (1  ±  x)n,  we  obtain 
a 

(l±8)Ll±liC-!L=J^±("-lX2»-l)^-...;    (6) 

n  ?r     2 !  ?r  3 ! 

Ex.      v'Tuoi)  =  ^1024  -  24  =  4  (l  -  -JL\   . 

V         128; 

Q 

Substitute  —  for  x  in  the  series 
128 

'  5      5  10      6  10  15 

The  result  of  this  substitution  to  six  figures  is  0.995268  ;  hence 
v/1000  =  4  x  0.995268  =  3.981072. 


arc  sin 


Art.  148]  USE   OF   SERIES   IN   COMPUTATION  307 

IV.    Computation  of  logarithms. 
The  series  for  log  (1  +  x),  that  is, 

x2  .  x3     x4  . 

converges  slowly,  and  is  therefore  not  well  adapted  for  computation. 
A  more  useful  series  is  derived  as  follows : 

/)**  />*>  rvA 

log(l-M)=*-|  +  |-|+.... 

Substituting  —  x  for  x,  we  have 

3/  fl»  £C 

By  subtraction  we  get 

log(l  +  x)-log(l-x)  =  logi±|  =  2^  +  |  +  J+...\ 

Let  us  take  x  positive  and  assume  x  = :  then 

2y  +  l 

l  +  a?  =  y  +  l 
1  —  x         y 
and  we  have 

^*+i>-^.+s[5^+i(^i+K^y+^ 

(7) 
From  this  series  log  (jf  + 1)  can  be  calculated  when  log  y  is 
known.     Thus,  since  log  1  =  0,  we  have 

i      o      Jl.ll   ,11,11  ,       \ 

=  0.693147..-. 

iog3=,og2+2(H!+!!+...) 

=  1.098612-... 
log  4  =  2  log  2  =  1.386294  •  •  -. 

log6  =  log4  +  2(|+|i+|i+...) 

=  1.609438.-. 
log  6  =  log  2  +  log  3  =  1.791759  .... 


308  INFINITE   SERIES  [Chap.  XV. 

Evidently  it  is  only  necessary  to  make  the  computation  in  the 
case  of  prime  numbers.  Logarithms  to  the  base  10  are  obtained 
from  the  natural  logarithms  by  means  of  the  following  relation : 

log10a  =  loge  a  x  — !— -  =  0.4342945-.  •  loge  a. 

10ge  10 

V.    Computation  of  trigonometric  functions. 
For  all  values  of  x,  we  have  the  series 


x      x3   ,  x5      x7   , 

sin  x  = h 

1!     3!     5!      7! 

r*d  /«4  /y,6 

-I  ms         .     Ms  *b         . 

cos*=l--  +  --_  + 


(8) 


These  series  converge  rapidly,  and  may  be  used  to  compute  the 
natural  sine  and  cosine  of  any  angle.  Necessarily  x  must  be  ex- 
pressed in  radians. 

Ex.  Find  the  sine  and  cosine  of  19°  30'  correct  to  five  figures. 

We  have  x  =  — —  it  =  .34034.     Substituting  this  value  for  x  in  the  two 
180  -° 

series,  we  get  sin  x  =  0.33381, 

cos  £  =  0.94264. 

149.  Approximation  formulas.  It  is  frequently  convenient  in 
computation  to  replace  a  function  by  another  of  approximately 
the  same  numerical  value  but  having  a  more  simple  form  or 
having  a  form  better  adapted  to  calculation.  This  substitution 
may  be  effected  in  many  cases  by  expanding  the  given  function 
and  taking  a  certain  number  of  the  first  terms  of  the  series. 

One  of  the  most  useful  of  these  approximation  formulas  is 
obtained  from  the  binomial  formula.  Thus  let  m  denote  a  small 
fraction,  and  expand  (1  ±  m)n.     The  result  is 

(l±m)w  =  l±wm  +  ^P^m2±  .... 

Z ! 

Since  m  is  small,  powers  higher  than  the  first  may  be  neglected, 
and  we  may  write  the  approximate  relation 

(l±m)n  =  l±nm.  (1) 


Akt.  149]  APPROXIMATION  FORMULAS  309 

An  important  special  case  is  that  in  which  n  =  \ ;  for  this  case 
we  have  approximately 

Vl  ±  m  =  1  ±  i  ra.  (2) 

From  (2)  we  have  the  more  general  formula 

V¥±l=a(t±£),  (3) 

as  may  be  easily  shown.     In  this  formula  b  is  small  in  compari- 
son with  a. 

Since  ew  =  l  +  m  +  -+-  +  •••,  we  have  when  ra  is  small  the 

£ .       o  ! 

approximate  relation 

em  =  1  +  m.  (4) 

Similarly,  taking   two   terms  of   the  series   for  sin  x,  cos  x,  and 
log(l  +  x),  we  obtain  the  approximate  formulas 

sin  ra  =  ra  (1  —  |  ra2)  ;  (5) 

cos  ra  =  1  —  \  ra2 ;  (6) 

log (1  -f- m)  =  m-\  ra2.  (7) 

When  ra  is  small  compared  with  a  the  following  approximate 
relations  are  readily  obtained. 

sin  (a  ±  ra)  =  sin  a  ±  ra  cos  a ;  (8) 

log(a  +  m)  =  log<.+5»-  "t;  (9) 

.        _J_  =  1-T»2  +  ^-  (10) 

a  ±  m     a      or     ar 

The  degree  of  error  due  to  the  neglected  terms  may  be  esti- 
mated by  taking  the  maximum  value  of  the  remainder  R.  For 
example  consider  the  approximation  (5).  If  only  two  terms  of 
the  series  are  used,  we  have 

sin  ra  =  ra  -  %-  +  ^-fr(m1), 

where  0  <  mx  <  ra.     Since  /  F(rax)  =  sin  m1  cannot  exceed  1,  the 
difference  between  sin  ra.  and  the  assumed  approximation  ra  —  — - 

5  5  6 

cannot  exceed  -^- ;  hence  i?  <£  ^-r  •       If  we  wish  to  restrict  the 


310  INFINITE    SERIES  [Chap.  XV. 

error  to  some  definite  limit  r,  we  have  only  to  write 

\B\<r 

and  solve  for  ra.  Thus  in  the  case  just  stated,  if  we  restrict  the 
error  to  one  unit  in  the  fourth  decimal  place,  we  have 

—  I  <  .0001, 
120 1 

whence  |  m  |  <  V-012,  or  |  m  |  <  .413.  Hence  for  angles  lying 
between  —  23°  40'  and  -4-  23°  40',  two  terms  of  the  series  give  the 
value  of  sin  m  correct  to  three  figures. 


Ex.  1.     The  relation  cos0  ='\/l  —  (- )   sin20  occurs  in  certain  problems 


relating  to  the  balancing  of  engines,  and  a  simpler  approximate  relation  is 

r      1 
desired.     Take  -  =  -  • 
I     6 

Putting  (-)   sin20=m,  we  have  for  the  value  of  m,   —  sin2  0,   and   this 
°  \l]  '36 

1  r2 

cannot  exceed  —     Hence  we  mav  use  (2),  and  write  cos*  =  1 sin20. 

36  ^  ;'  T  2l2 

The  expression  for  B  is 

m 

'2 


/"(mi)  ^-  ==  - — -  (Since  f(m)  =  Vl  -  m) , 

(1  —  nny 


The  maximum  value  of  m  is  fa  and  since  0  <  mi  <  w,  the  error  of  the  approxi- 

1  (  1  V2 
mation  cannot  exceed  -  i3J^L_  ~  0.0001. 

Ex.  2.     In  the  theory  of  centrifugal  fans  the  pressure  ratio  is  given  by 

2?  =  ek,  where  po  and  px  denote  respectively  the  pressure  of  the  air  entering 
Pi 

the  fan  and  that  of  the  air  leaving  it.     The  exponent  k  is  a  constant  depend- 
ing upon  the  speed  of  the  fan  and  is  small.     Taking  k  =  0.04,  we  have 

approximately  ^i  =  1  +  k  =  1.04.     To  determine  the  error,  we  have  for  the 

Pi 

k2  k2 

remainder  f"(ki)  --  =  e*i  — ,  the  maximum  value  of  which  is 

go.o4(°-04>)2-  Q.00083. 

2 

Ex.  3.     In  certain  problems  in  surveying  the  relation  between  a  circular 
arc  and  its  chord  is  required.     Let  s  denote  the  length  of  the  arc,  r  the  radius, 

and  C  the  chord,  Fig.  101.     We  have  s  =  ra,  and  C=2rsin  — •     If  a  is 


Arts.  149,  150] 


MAXIMA   AND   MINIMA 


311 


small  the   approximation  (5)  may  be  used.     An  approximation  to   G  is 
therefore 

°=2'iH{f)'] 

=  ra  —  ^  ras. 

Hence,  s-  C  =  j»j ra3, 

where  a  is  expressed  in  radians.     If  a  is  taken  in  de- 
grees, the  formula  becomes 

s-  G 


ras 


4514180 

The  error  of  the  approximations  cannot  exceed 

ra5 


2r 


120 \2  ) 


1920 


Fig.  101. 


EXERCISES 

1.  Calculate  sin  15°  and  cos  15°  to  five  decimal  places. 

2.  From  the  series  for  tana;  calculate  tan  12°. 

3.  Find  ^1334. 

4.  From  the  logarithms  given  in  Art.  148  calculate  log  31,  also  log  73. 

5.  Using  the  approximate  formula  (10)  calculate  the  reciprocal  of  102 ; 
of  97. 

6.  Find  the  greatest  value  of  m  that  will  permit  the  approximation 

(1  +  ra)4  =  1  +  4  ra 
with  a  maximum  error  of  1  in  1000. 

7.  Within  what  limits  will  three  terms  of  the  series  for  cos  x  give  an  error 
not  exceeding  2  units  in  the  6th  decimal  place  ? 

8.  Investigate  the  limits  of  accuracy  of  the  formula 

sin  (a  +  ra)  =  sin  a  +  m  cos  a. 

9.  Find  the  length  of  the  chord  of  an  arc  of  radius  200  feet  subtending 
an  angle  of  3° :  (a)  by  trigonometric  methods  ;  (6)  by  the  approximation 
formula.     Compare  the  results  and  find  the  relative  error  of  the  approximation. 

10.  Derive  an  approximation  formula  for  tan  x  and  show  the  maximum 
error  involved. 

11.  Given  log  5  =  1.6094,  find  log  5.01  and  log  5.02. 

ISO.  Maxima  and  minima  of  functions  of  a  single  variable. 
Taylor's  expansion  of  a  function  gives  a  convenient  method  of 
developing  the  condition  for  maxima  and  minima.  This  method 
is  particularly  valuable  where  several  of  the  derived  functions 


312  INFINITE   SERIES  [Chap.  XV. 

vanish  for  the  critical  value.     The  condition  for  a  maximum  or 
minimum  of  a  function  in  such  cases  may  be  stated  as  follows : 

Theorem.  The  function  f  (x)  has  a  maximum  (or  minimum)  for 
x=a,  if  the  first  one  of  the  derived  functions  f'(x))  f"(x)  •••  that 
does  not  vanish  for  x  =  a,  is  of  even  order  and  negative  (or  positive). 

We  have  from  Taylor's  theorem, 

f(X+h)=f(X)+f\X)h+l^h^£^w+ ...  +fn(x+eh)h% 

and 

fix  -  h)  =f(x)  -f  (x)  h + tM  n*  _  f!!M  &»+...+  £fe± eh)  h\ 

Replacing  x  by  a,  we  have,  after  transposing  the  first  term  of  the 
second  member  of  the  identity, 

/(a  +  ft)  -  /(«)  =/'  (a)  ft  +  -f^fl  h°  +  £^2)  ft»  +  ..  . 

+  /"(«  +  flft)/t,, 

n ! 

f"(a)h2     f'"(a)h* 
2!  3! 

I  (     ±yfn(a~^~    l^hn.  (2) 


f(a-  ft)  -  f(a)  =  -  /'  (a)  ft  +  LJ£1  ft«  _  -£_p  ft"  + 


If  for  x=a  the  given  function  has  a  maximum,  then /(a)  must 
exceed  the  value  of  the  function  for  all  values  of  the  variable  in 
the  neighborhood  of  a ;  in  other  words,  the  left-hand  member  of 
both  (1)  and  (2)  must  be  negative  for  all  values  of  h  sufficiently 
small.  The  value  of  h  can  be  taken  so  small  that  hf'(a)  will  be 
numerically  greater  than  the  sum  of  the  remaining  terms  of  the 
second  member.  However  f(a  -f  h)  —  f(a)  and  f(a  —  h)  —  f(a) 
cannot  both.be  negative  unless  f(a)—0  and  f"(a)  is  negative, 
assuming  that /"(a)  does  not  vanish.  It  may,  however,  happen 
that  both/'(«)  and  f"  (a)  vanish.  In  this  case,  in  order  to  have 
f(a  +  h)  —  f(a)  and  f  (a  —  h)  —  f  (a)  negative,  f'"(a)  must  vanish 
and/IV(a)  must  be  negative.  In  general,  in  order  that/(#)  shall 
have  a  maximum  value  for  x  =  a,  the  first  derivative  that  does 
not  vanish  must  be  of  even  order  and  negative. 

In  order  that  f(x)  shall  have  a  minimum  for  x  =  a,  the  two 
expressions  f(a  -f  h)  —  f(a)  and  f(a  —  h)  —  f(a)  must  be  positive. 


Art.  151]  INDETERMINATE   FORMS  313 

This  requires  that  the  first  derivative  that  does  not  vanish  shall 
be  even  and  positive.  The  argument  is  similar  to  that  given 
above  and  is  left  to  the  student. 

Ex.     Examine  the  function  2  cos  x  +  ex  +  e~x  for  maxima  and  minima. 
We  have  f(x)  =  2  cos  x  +  ex  +  er-  ; 

f'(x)  =  —  2  sin  x  +  ex  —  e~x. 
For  x  =  0,  f'{x)  =  0  ;  hence  x  =  0  is  a  critical  value.     We  have  further 
f"(x)  =  -  2  cos  x  +  ex  +  «-*,  whence  /"(0)  =  0, 
/'"(«)  =  2sina;  +  e*  -  e-z,  whence /'"(0)  =  0, 
/iv(x)  =  2  cosic  +  eK  +  e~x,  whence  /IV(0)  =  4. 
Since  the  fourth  derivative  is  positive,  and  is  the  first  that  does  not  vanish,  it 
follows  that  f(x)  is  a  minimum  for  x  —  0. 

EXERCISES 

Examine  for  maxima  and  minima  the  following  functions. 
1.    tan2  x  —  2  tan  x.  2.    ex  —  e~x  —  2  sin  x. 

3.   sinx(l  +  cosse).  4.  peax  +  qe-™. 

5.    icsinx.  6.   3  cos  0  +  tan2  0. 

7.   cos  z  —  log  cos  x.  8.   x_men,a;. 

9.    ex  —  e~x  —  a;2.  10.   cos  x(2  —  cos  x). 

151.    Evaluation   of   indeterminate  forms.     Algebraic   methods 
of  evaluating  certain  indeterminate  forms   were    shown   in   the 

examples  of  Art.  15.     For  the  form  -,  to  which  all  other  forms 

may  be  reduced,   the  differential   calculus   furnishes   a   general 

method  of  evaluation,  which  is  developed  as  follows : 

fix) 
Let  the  given  function  be  of  the  form  J  \  ' ,  which  reduces  to  the 

<f>(x) 

0  f  (x) 

form  -  for  x  =  a.     The  value  of  the  limit     L   *-*-£  is  required. 
0  x±a<f>(x) 

Expanding  each  term  of  this  fraction   by  Taylor's  theorem,  we 

have 

,,  ,     f(a)+f\a){x-a)+-f-^-(x-af+  -  +€^L(x-ay 
/W_ «■ JJj (1) 


314  INFINITE   SERIES  [Chap.  XV. 

By  hypothesis, /(a)  and  <£(a)  are  each  equal  to  zero.     The  above 
relation  therefore  reduces  to 

f(x)     f'^x-a)+-t^li-x-a^+  -  +^(*-«)" 
^%'(a)(z-a>+«(*-a)s+---+M  (*-<»)• ' 


(2) 


Dividing  both  terms  of  this  fraction  by  (x  —  a)  and  passing  to 
the  limit,  we  have  ^  f(»)  =fjg) 

x±a4>fr)      *'(*)' 

If  /'(a)  =  0  and  cf>\a)=^0,  this  limit  reduces  to  zero;  if  /(a)  =£0 
and  <f>'(a)  =  0,  it  becomes  infinite. 

If /'(a)  =  0  and  <f>'(a)  —  0,  the  limiting  value  of  the  given  func- 
tion can  be  obtained  by  dividing  the  terms  of  the  expanded  form 
of  the  fraction  (x  —  a)2  and  then  passing  to  the  limit.     The  result 

is  L  ^M=/M.  a) 

x  =  a<f>(x)      <f>"(a)  V  ' 

Similarly,  if /'(a)  and  <£"(«)  are  both  zero,  we  divide  by  (x  —  a)3 
and  again  take  the  limit,  and  so  on. 

We  may  therefore  state  the  general  law  of  procedure  as  follows : 

To  evaluate  the  indeterminate  form  -,  differentiate  the  numera- 
tor and  the  denominator  of  the  given  fraction  and  substitute  the  criti- 
cal value  of  the  variable  in  the  result. 

The  function  «^4  may  ^so  assume  the  indeterminate  form  - 
<f>(x)  0 

when  x  becomes  infinite.     The  limiting  value  may  still  be  found 
by  considering  J  }  '  \  for,  we  have  upon  putting  x  =  -, 

L    ./M=    L    —gjj**    L       -M         T     t!iA      (5) 

Form  3£ .     When  the  function  £iw2.  takes  the  form  ^- ,  it  can 

00  <f>(x)  °° 


Art.  151]     EVALUATION   OF  INDETERMINATE   FORMS  315 

1 

be  reduced  to  the  form  -,  by  writing  it  in  the  form  Z-j-J.     This 

m 

form  can  then  be  evaluated  according  to  the  law  just  stated. 

It  may  be  shown  as  in  the  case  of  the  form  -  that  if  U&  has 
J  0  cf>'(x) 

a  limit  as  x  approaches  a  definite  number  or  becomes  infinite,  then 

fy*f   converges  to  the  same  limit.*     This  principle  often  affords  a 
<f>(x) 

convenient  method  of  evaluating  this  indeterminate  form  ;  for  we 

need  only  to  differentiate  the  numerator  and  the  denominator  and 

then  pass  to  the  required  limit. 

Form  od  —  ao .     When  a  function  takes  the  indeterminate  form 

oo  —  oo  it  may  be  reduced  to  the  fundamental  form  -  by  writing 
it  as  follows : 

,w-^44,t^i.       (6) 

J(X)        0(0)        f{x)  •  *(0) 

Often,  however,  a  simpler  transformation  will  reduce  the  function 

to  one  of  the  forms  -,  —  •     No  general  rule  can  be  given,  but 

that  transformation  should  be  selected  which  gives  the  simplest 
form. 

Form  0  x  ao.    When  a  function  fix)  •  cf>(x)  takes  the  form  0  X  oo 

for  x  —  a,  it  may  be  reduced  to  the  type  -  by  writing  it  in  the 
form 

m.m„f!gL  (7) 

<t>(x) 

Forms  0«,  oo°,  1*.     The  indeterminate  forms  0°,  oc°,  1"  arise  from 
a  function  of  the  form  [/(a;)]*(x).     This  function  may  be  reduced 

to  the  type  form  -  as  follows : 


*  See  Pierpont's  Theory  of  Functions,  Vol.  I.,  p.  305.   The  special  student 
of  mathematics  would  do  well  to  read  Arts.  455-459  in  the  same  volume. 


316 


INFINITE   SERIES 


[Chap.  XV. 


Let  y  =  [mTx\ 

whence  logy  =  <f>(x)  •  log  [/(#)]•  (8) 

Since,  for  each  of  the  given  forms,  (8)  takes  the  form  Oxoo,  the 
solution  is  effected  by  (7). 

n      •■     <n—_i    ~  l    tan  x  —  sin  x    £  n 

Ex.  1.   Evaluate ,  for  x  =  0. 


We  have 


f'(x)  _  sec'2  x  —  cos  a;" 


Hence,      L 

(T  =  0 


f"(x)  _  2  sec2  x  tan  x  +  sin  an       _  0 
<p"(x)~  6x  J*=o~0 

f"'(r)  _  4  sec2  %  tan2  x  +  2  sec4  x  +  cos  % 
0"'(a;)  ~  6 

tan  x  —  sin  x  _  1 
x*  ~2* 


J*=o      2 


Ex.  2.    Evaluate  sec  -  irx  •  log  -  for  x  =  1. 

2  x 

This  function  takes  the  form  0  x  oo  ;  hence,  we  write  it  in  the  form 

1 


log- 


ic* 

X  X 


sec  \  irx 
0 


COS  \  irx 


which  for  x  =  1  takes  the  form  -  •     Differentiating  the  numerator  and  the 
denominator,  we  get 


—  log- 
dx       x 

~l         1 
—  cos  -  irx 
dx        2 


1  .     1 
ir  sin  -  irx 

2  2 


Ex.  3.  Evaluate  xl~x  for  x  '==  1. 

Let  2/  =  x1-a!;  then   logv  = log  x  =   °^ x     which  has   the  form  -. 

1  —  x  1  —  x  0 


Hence,  by  the  general  rule 


L  logy 

5=1 


-flogx 
dx 


whence 


(1-x) 


1  " 

X 

- 1 


h 


L  y  =  ± 

x=i        e 


Arts.  151,  152]  SINGULAR   POINTS  317 

EXERCISES 

Evaluate  the  following  indeterminate  forms  : 

Xs  +  2  x2  -  x  -  2" 


1. 

x2  +  4  x  -  21 

x3  -  3  x2  -  4  x  +  12 

3. 

1  —  COS  X~| 

x2       _L=o ' 

5. 

sec  x  —  tan  x]    „.. 

x~2 

7 

tan  0  —  0~| 
fl-sinflj^o 

9. 

1     *1   • 

sin2  x      x2Jx=o 

11. 

e*  sin  - 

#Jx— oo 

13. 

sin  x  log  cot  x]x=o. 

15. 

zx]*=o. 

17. 

i 

(e*  +  a:)z]i=0. 

19 

tan  x  —  x"j 

£3          Jx=0 

?1 

ex  sec  x  —  l~j 

*             Jx=0 

91 

X                    1       *1 

2     X*  +  2  ag  -  x  —  21 

x2  +  10  x  +  16   Ja 

4.    I  (a* -&*)"]      • 

x  Jx=0 


6. 


logg 


L 


Vl-x- 

8.         **-* 1 

1  -x  +  logxJx=i 

10.    x(a*-  l)]x=oo. 

12.    ?-cot?l      • 

x  2_Uo 

14.   (sin0)tan0]       . 
e~2 
l 
16.    (l  +  x2)-]x=0. 


18.    sec  2  0  cos  5  0]    «, 


20. 

arc  tan  x  —  5 

■1   • 

Jx=0 

X3 

22, 

irX 

sec  — 

2 

log(l-x)J 

x=l 

24. 

x            1 
x  —  1     log  i 

Ux=l 

152.    Analytic  condition  for  a  singular  point.     In  Art.  89  atten- 
tion was  called  to  certain  points  of  plane  curves,  called  singular 

points,   where  the   derivative  -^  has   not   a   single  determinate 

ctx 

value.     At  such  a  point  —  must  have  therefore  the  indetermi- 

dx 

nate  form  -•     Having  now  a  general  method  of  evaluating  this 

indeterminate  form,  we  can  deauce  an  analytic  method  of  deal- 
ing with  singular  points. 


818  INFINITE   SERIES  [Chap.  XV. 

Let  the  equation  of  the  curve,  written  in  the  implicit  form  and 
without  radicals,  be  -•      v  _  q  ^ . 

Then  for  a  singular  point,  we  have 

^__^-A  /ox 

dx~      d£     0'  ™ 

By 

that  is,  in  addition  to  f(x,  y)  =  0,  we  must  have  -J-  =  0,  -^-  =  0. 

Solving  these  equations  simultaneously,  we  find  the  coordinates 
of  points  at  which  singularities  occur.  Having  found  such 
points  we  may  determine  the  character  of  the  singularity  at 
any  one  of  the  points  by  evaluating  the  indeterminate  expres- 
sion for  -^  by  the  methods  already  developed.  The  following 
cfx 

examples  will  serve  to  illustrate  the  method  of  procedure. 
Ex.  1.    Examine  for  singular  points  the  curve 

f(x,  y)  =  4  x3  -  12  x2  +  10  x  +  xy  +  11  y  -  3  y'2  -  14  =  0. 
We  have         •>  *<■ 

°JL  =  12  x2  -  24  x  4-  10  +  y,  ^L  =  x  +  ll-6y. 
dx  dy 

The  values  x  =  1,  y  =  2  satisfy  the  equations 

f(x,y)  =  0,  |£=0,  &  =  0; 
ox  oty 

hence  the  point  (1,  2)  is  a  singular  point.    The  character  of  the  singularity 

dy 
dx 


can  be  determined  by  evaluating  the  indeterminate  form  -^-     For  con- 


venience putting  -^  =  p,  we  have 
dx 

_  _dx_      12  x2  -  24  x  4-  10  +  y 
P_      fl£~  x  +  ll-6y 

The  right-hand  member  of  this  equation  takes  the  form  -  for  x  =  1,  y  =  2  ; 

hence  to  determine  its  value  at  this  point,  we  differentiate  both  the  numera- 
tor and  the  denominator  with  respect  to  x  and  have 


24  x  -  24  4- 
l-6p 


£l 


Art.  152]  SINGULAR  POINTS  319 

or  p  =    ~p    , 

*      1-6/ 

whence  2p  —  6p2  =  0, 

p  =  0,  or  i. 

Therefore  at  the  point  in  question,  there  are  two  distinct  tangents  to  the 
curve  and  consequently  two  branches  of  the  curve  pass  through  that  point 
and  the  singularity  is  a  double  point.  The  slopes  of  the  two  tangents  are  0 
and  ^,  respectively. 

Ex.  2.   Examine  for  singular  points  the  curve  x4  +  xsy  —  4  x2y  -f  y8  =  0. 
We  have  for  the  partial  derivatives 

&  =  4 a*  +  3 x2y  -  8 xy,     ^  =  x3  -  4  x2  +  3  */2. 
ox  dy 

The  values  x  =  0,  x  =  0  satisfy  the  equations  /(x,  y)  =  0,    %£  =  0,    zL  =  0  ; 

dx  dy 

hence  the  origin  is  a  singular  point.  Putting  SlU  _  p^  we  have  for  the  point 
(0,  0)  dx 

4  Xs  +  3  x2y  -  8  xyl      =0 
it-s  _  4  X2  +  3  y2    JM  =  0? 

whence  differentiating  numerator  and  denominator  with  respect  to  x, 
we  obtain 

_  12  x2  +  6  xy  +  3  x2p  -  8  y  -  8  ay  "I 

3x2-8x  +  6*/.p  J  o,o 

For  x  =  0,  y  =  0  the  second   member  again  takes  the  form  2  ;   hence  dif- 

0 
ferentiating  numerator  and  denominator  a  second  time,  we  have 

_  24  x  +  6  y  4-  6  xp  +  6  xp  —  8 p  —  8  p~|      =     8j? 


Jo,  o 


6x-8  +  6p2  J00     3p2-4 

Therefore  jt? (3 p2—  4)  =  8p,  andp  =  0,  2,  —2.     The  origin  is  a  triple  point, 
and  the  three  tangents  have  respectively  the  slopes  0,  2,  and  —  2. 

EXERCISES 

By  the  general  method  of  this  article  examine  the  following  curves  for 
singular  points.  Additional  exercises  are  furnished  by  the  examples  of 
Art.  89. 

1.  x4  -  4  xhf  -f  ys  =  0. 

2.  x3  -  3  x2  -  3  y2  +  y3  =  0. 

3.  x5  -  4  x4  +  ys  -  4x2  y  =  0. 

4.  x3  +  2  x2  -  4  x«/  +  2  y2  =  0. 

5.  x4  -  x2y  +  y8  -  6  y2  +  2  x  -  12  y  -  8  =  0. 

6.  ay  =  (x  -  &)*. 


320  INFINITE   SERIES  [Chap.  XV. 

7.  Trace   the   curve    x4  +  x8y  —  4  xHj  -f-  yz  =  0,    discussed   in   illustrative 
Ex.2. 

Suggestion  :   Examine  for  asymptotes,  then  put  y  =  rax  and  find  values 
of  x  for  assumed  values  of  the  slope  m. 

8.  Trace  the  curve  of  Ex.  1.  9.    Trace  the  curve  of  Ex.  6. 

MISCELLANEOUS    EXERCISES 

1.  Test  for  convergence  the  following  series,  and  determine  the  interval 
of  convergence  : 

?a\  i_i_W*\  .    1  l-4/x\2       1  1-4-7 /xy, 

»l+!+ii7+ii5+--;- 

(c)  1  +xcosa  +  —  cos  2  a  H — cos  3  a  +  •••. 

2!  3! 

(d)  cosx  +  -cos2x  +  -cos3x  +  •  ••. 

2  3 

Expand  the  following  functions. 

x  0 

2.  tan0.        3.    e*cosx.  4.   cosnx.         5.    .        6.   log  sec2-. 

e*-l  2 

l 

7.  Show  that  (1  +  *)«  =  e  (1  -  |  x  +  §£  x2  -  T\  x3  +  •.•)• 

Suggestion:  Let  u  —  (1  +  x)x,  whence  log  w  —  _2S_L_±_^2.     Make  use  of 

the  series  already  found  for  log  (1  +  x)  to   determine  the   successive  de- 
rivatives. 

8.  From  the  expansion  for  log  (n  +  h)  and  log  (n  -J- 1),  derive  the  approxi- 
mate rule  of  proportional  parts,  viz. : 

log  (n  +h)  —  log n  _h 

log  (n  +  1)  —  log  n      1 ' 
From  this  rule  find  log  7.523,  knowing  that  log  7.52  =  2.0176  and  log  7.53  = 
2.0189. 

9.  Develop  f(h±K)   in    powers   of  h,    and    determine    the   values    of 
/(x)  =x2(16  -  x)  for  the  following  values  of  x  :  4.7,  4.8,  4.9,  5,  5.1,  5.2,  5.3. 

10.  Show  by  development  in  series  that  exS/~~^  =  cosx  +  V—  lsinx,  and 
e-xV-\  =  cos  x  —  V—  1  sin  x. 

11.  Develop  the  function  y  =  arcsina?  in  series. 

VI -x2 


Suggestion  :    Multiply  by  Vl  -  x2  and  differentiate.     The  resulting  equa- 
tion is  (1  —  x2)  =£  —  xy  =  1.     Now  assume  y  =  A  +  Bx  +  Cx2  +  •••  and  deter- 
dx 

mine  the  coefficients. 


Art.  152]  MISCELLANEOUS   EXERCISES  321 

12.  Derive  the  approximate  formula  tan  (d  -f  w)  =  tan  6  +  m  sec2  6. 

13.  The  strength  of  an  electric  current  as  shown  by  a  tangent  galva- 
nometer is  given  by  i  =  C  tan  <f>,  where  C  is  a  constant  and  <p  is  the  deflection 
of  the  needle.     Show  that  an  error  m  in  the  reading  of  the  angle  gives  a 

2  in 

relative  error  of in  the  current. 

sin  2  0 

Evaluate  the  following  expressions  by  the  use  of  series  : 

14.  *=ii    .      is.  *-sin*i    .      is.  +-■**]   : 

x     J*=o  xs      J*=o  sinx    _|x=o 

Calculate  the  following. 

17.  Sin  10°  and  cos  10°  to  five  figures. 

18.  Logarithms  (natural)  of  17,  31,  61,  correct  to  four  places. 


20.  Prove  that  the  expansion  of  an  even  function  of  x,  that  is,  one  for  which 
f(x)=f{—  x),  will  contain  only  even  powers  of  a*,  while  if  f(x)  =  —  /(_  x) 
the  expansion  will  contain  only  odd  powers  of  x.  Illustrate  by  several 
functions. 

Making  use  of  the  known  series  for  ex,  log  (1  +  x),  sinx,  etc.,  derive  series 
for  the  following  functions. 

ex 


21.   — 22.    ex log (1  +  x) .        23.    Vl±sin2x.         24.    cos2x. 

25.  By  substituting  mx  for  x  in  Ex.  10,  prove   DeMoivre's  theorem, 
namely,  (cos  x  _|_  y/_  \  sm  xyi  —  cosmx  +  ^/_  \  s[n  mx# 

26.  By  means  of  the  exponential  series  and  the  identity 

show  that  2  sin  x  cos  x  =  sin  2  x. 

27.  Denoting  by  c  the  chord  of  the  half  arc.  Fig.  101,  derive  Huygen's 
approximation  to  the  length  of  a  circular  arc,  viz. :  s  = • 


Suggestion.     Expand  sin-  and  sin—  in  series  and  combine  the  results. 

28.  Referring  to  Fig.  101,  deduce  an  approximate  formula  f or  s  —  C  in 
terms  of  the  chord  C  and  the  dimension  h. 

29.  Examine  for  singular  points  and  asymptotes  the  curve 

x5  —  5  ax2y2  +  y~°  =  0, 
and  trace  the  curve. 

30.  Examine  for  singular  points  and  asymptotes  the  curve 
Trace  the  curve. 


ANSWERS 


The  answers  to  some  of  the  problems  have  been  purposely  omitted. 

Art.  8.     Page  11. 

1.    _  26  ;   -  14  ;   -  110.  3.    1  ;  \  V2  ;  0  ;  -  1. 


6.    3  -  V^+4  ;  (3  -  Vx)2  +  4  ;  (y2  +  4)2  +  4  ;  3  -  V3  -  Vx. 
„        _  _  1  14.    cos  8 ;  sin  0. 

x  15.   sec0. 

12.    sin  (x  ±y).  16.    tan(x  —  y). 

Art.  12.     Page  17. 
2.    £  =  2,  £  =  3.     4.   (a)  Discontinuous  at  £=0.     (6)  Discontinuous  at  x=0. 

Art.  15.     Pages  24,  25. 

1.  -f.  3.    0.  5.    48.  11.    -f;  0. 

2.  i.  4.    1.  6.    5  a4.  12.   1. 

Miscellaneous  Exercises.     Page  25. 
1.    (a)  (2  » -3)  Ax  +  (As)*.     (5)0.31.     2.  (6)  8  a*2.      12.   (a)  J.     (6)  J. 

Art.  17.     Page  30. 


1. 

4x3. 

5.    3x2-l. 

9.    a  +  gt. 

2. 

2x-4. 

2a 
x3  ' 

6                X 

in            & 

3. 

(*-l)a 

7.  3(x-a)2. 

8.  flft. 

10.    a  —  —  • 

02 
11                       ^ 

4. 

O  -  &)2 

Art.  28.     Pages 

37, 

38. 

1. 

6x-4. 

2. 

12 

x2-4x.         3.    3  a:'2- 

-  10 

X. 

4.    3x2-2x- 

-2. 

5.  x2(6x3  +  5x2-8x-6).                    7     6-x  8     __2# 

6.  6x2-8x  +  3.                                                x3  '        x3' 

9              ma                        II'  1  —  4  13     2(2x2-3x-4) 

(ax  +  &)2'                    '          Vs'  (x2  +  2)2 

in      «d-frg                        12         5«8+1)  M-  -x2  +  12x  +  19 

'      (CX  +  d)2                                                     («2-l)2  '        (x2  +  4a;_5)2 

323 


324  ANSWERS 


Art.  30.     Page  40. 

1.    §Vx.  2.    2x  +  x~2.  n.    Dtp=  -~-^- 

t2       ts 


3. 


U 


\Vx     Vx3/ 


2\Vx      Vx3/  12.    DdP=2  0+ 


k 


4.  8(2  x-5)3.  2v^ 

5.  4(x-2)(x2-4x  +  3).  M-   x*-\l  -  x)*-i(p  -  px-qx), 

6.  3x(x2-a2)i  15.    x2(4  x<2  ~  15)  • 
o,                _ 2  Vx2  -  5 

7.  ~(l+x2)  * 


8 


16. 


8.   -t(4s  +  8)~i.  8x*(x2-a2)* 

2a2x 


18.   y  =  31og?. 


^2  _  a2)i(a;2  +  a2)l 


10.    «/«+26  +  3cV  19.   x  =  aarccos^^±V2 
\x2      x3       x4 )  a 

Art.  31.  Page  42. 

1.   2X(z2  +  5)-l  9> 


ay  —  j/ 


3(x8-2x  +  5)J 


10.   8 


Vx2  —  2  x  +  5  2 

3.    -5x(a2-x2)i  11.        /~g2     - 

2  0*(1  +  A2)* 


(5x2  +  6x-12)J^±4' 

'x2  —  4 


12.    -0(1  + 


4.    5x(a2-x2)  * 

6    2x*(x2-2a2)  '        2Vx2-a2 

3(x2-a2)^  u  x3 

7.  (2x-f  a)(3x2-4ax)"l  (x2  +  1)^ 

8.  m  +  n  +  2x  15.   x(2  <*  ~  gQ . 
2V(x  +  m)(x  +  w)  (a2-x2)$ 


Art.  32.     Page  44. 


3.   -g 4.   JJ£.     6.   \{ev-fvr 


3(*_4)*  (l-*2)2  2V^T&  >2s 

Art.  33.     Pages  44,  45. 

1.  e,_i.        *»£=*-'.  s.  _L.  •.»=».  jA, 


ANSWERS  325 

Miscellaneous  Exercises.     Pages  45,  46. 

1.   20^  +  9a;2_8.  i*  x2-6x-5 


-8). 
a*). 

-3^ 

4yu 

sV2 

.  37. 

3.    0 

2(x-3)%Vtf~+~5 

2.  x(\bx*  +  \2x2-\2x- 

3.  *af*(4a*-a*)(**- 

1d     -(2  +  aOVl-s2 
(1  +  x)2 

15.   2*+      2*3     . 

4     2x-5 
(a-  -  2)3 

Vx4-  1 
1f?    8x4-9a¥  +  3aa* 

5.    -     *      . 

3(x  +  a)(a?-a2)$ 

VI  +  x2 

6.    § 

x*Vx2  —  1 

7             ! 

17.  2g~2a^~^.. 

Vx2  -  i 

18.  4x-2(2^2  +  1). 

Vx2  +  1 

4Va+v^ 

ifl    2a            c 

(1  +  x)«+i 

V3        (0  -  6)2 

q         nx  +  mx  +  aw 

c  +pma 

xn+1(a  +  x)w+i 

22.  arc  tan  (±  f). 

1n     n(x  +  Vx*-l)n 
Vx2-  1 

Q?         10x(2-x2) 
(x4-  10x2+  10)^ 

11.  Vx  +  Vl  +  x2 

2Vl+x2 

12.  * 

24.  a    • 

6  +  v 

25.  _£-  -<?. 

v2        p2 

(x  +  1)  Vx2  -  1 
?7    fi(l-2fix-  X*)     1     l 
(z  +  M)2         '  M' 

26.    (6)   1.0025025. 

;  -At+Vl  +  M2. 

28-    0;    ±f;    q=|.              29. 

ac                  go           ^ 

Art 
2-    (a)  J.     (6)  J. 

'as -2  s2                    (ax  +  6)2 

Page  55. 

L50925. 

Art.  38.     Page  56. 

1.  3x-8y=_38;8x+3y=-4. 

2.  9x  +  4y  =  72;4x-9y=-65. 

3.  2  x  —  ij  =  a;  x  +  2y  =  3a. 


326  ANSWERS 

4.  3x  +  4t/  =  50;4x-3y  =  25. 

5.  8  x  +  5 V21  y  =  100  ;  5V21  x  -  8  y  =  J£  V21. 

6.  9x-y  =6;  x+9y  =  ll0. 

7.  17  x  -  4  ?/  =  20  ;  8  x  +  34  y  =  135. 


8. 

3  yi2y  =  axi(2  x  -  x{)  ;  y  -y1=  -  -&-  (x  -  Xi). 

—  #Xi 

9. 

Xi3                                                  ?/l3 

10. 

§#!«-*  +  xxi"-1  =  an;  y-ij!=  t^J"  (x  -  Xi). 

11. 

xjt1_yy1=l                _at]Ll{        ^ 
a2       62         '                      &'2X! v           ; 

Art.  39.     Page  58. 

1.    f  VT3  ;  2| ;  0.  3.    4  ;  4  ;  4  V2  ;  4V2. 

4.  *i;  tf.  Vx7+^;  '^^+li2.  5.   !;f;f;|. 

Xi  Xi 

6. ^ ;  SwiCxi2-!);   ^ V9  Xi4  -  18  xx2  +  10  j 

3(xi2-l)'      yK  J'  3(xx2-l) 

?/iV9xi4-  18xi2+  10. 

7.  ^;   ^;  |VI8;  fVH.  8.    |j  2«;   ^V5;  «V5. 

V*  '  xt 


Art.  41.     Page  61. 


2.    2aV7i;  -2_  .  3.    -^^/l - 


1_ 
fcI  y^'4** 


4.   a0Vl  +  0s  ;  ad2 ;  aVl  +  02 ;  a. 
6>   a^vp?.  «*        ^^ 


6    (g  +  i)V(^  +  g)2  +  i;  (#+'1)t.  V(02  +  0)2  +  i,  lu 


7    pV4p4+  («  +  2  60)2.       2p«         V4  p4  +  (a  +  2  60)2 ,    aJ-_260# 
a  +  2  60  '  a  +  2 60 '  2  p  2  p 

8.    0;    *;  02  +  0;  2  0  a  +  K. 
2'  a +  260 


ANSWERS  327 

Art.  43.     Page  64. 

1.   (a)  v  =  Vf>  —  gh  \  a  =  -g.  5.  (a)  32  rad./sec.;  -  32  rad./sec.2. 

(6)  236.6  ft./sec.;  -32.2ft./sec.2.  (6)  3£  sec. 

(c)   -  86.4  ft./sec.  6.  w  =  a  -  3  bt2  ;  a  =  -Gbt. 

4.  13  rad. /sec. ;  2  rad. /sec.2.  8.  w  =  6  +  2  c< ;  «  =  2c. 

Art.  44.    Page  66. 
1.  0.54084;  0.23440;  0.23848.  2.   -0.00006704;  0.00026816. 

Miscellaneous  Exercises.    Pages  67,  68. 

1.  (a)  o ;  —  £.     (6)  At  x  =  0  and  as  =  f  a.      (c)  At  x  =  —  f  a  and  x  =  2  a. 

2.  ±450.  3.  15x  +  2y  =  60. 

9.   (a)  p  =  kd  +  c.     (b)  6  +  -  =  ft     11.  0.50925  ;  0.51040. 

Art.  48.     Page  77. 
1.  d|f  = 2eZx     .  8    dy_ 


Cx~5)2  (2«x-x2)* 

(a2  +  x2)Va*-^  _    wrtfc 


3.  dy=  (2x-5a)dx,  *  V2^Tl 

15x*(x-a)* 

5.  <fy  =  bmn(a  +  6x")m    x"    dx. 

l  +  3x^  11.  dy 


6.  <fy  = f dx.  x2Vx2  +  a2 

2  VI  +  x 

7.  dy  =_.(«*-«*)*«.  12.  <fr  =      8^2> 

2x*  3(x2+l)* 

Art.  49.    Page  78. 

3x2-y2  4    4x«-3x2y  +  2xy2  +  2y8 

2xy     '  '          x»-2x2y-6xy2 

tPx  5        ax  +  hy  +  g_ 

3.   -3x?/i  6.  |. 


328  ANSWERS 

I"  ~*5  *'  10.   (a)  -Mj  (6)  F8-«H2«6 

»•   1.  mv                vs(b  -  v) 

9.   (a)  Subtan.  =  4/>2v/0-2P0;  11     2y(l  +  x)  +  3x2  +  y2 

Subn.= fi >*!-*>-(*+«)»■ 

4,v*-2*  i2.  -y  +6 . 

(6)  Subtan.  =  1  -  2  P0 ;  2*y 

Subn.=^^.  13.   *■   ^~2^. 

l-2Pd  x      Vx-2Vy 

Art.  50.    Pages  80,  81. 

1.  Vi  1^34.  7.  _JU_. 

2.  12;   -9.  V^Z 

3.  (a)  6  ft./sec. ;  (6)  10  ft. /sec.  8"  0"25  Per  cent  aPProx' 

4.  (a)  6  =  ap;  (6)  ,2  =  **.  9-   ^f=  8  ^- 

5.  2  7r  cu.  in. /sec.  70 

«    03      •  ,u  n-   —  ft./sec. 

6.  3|  mi./hr.  tt 

Miscellaneous  Exercises.     Pages  81,  82. 

1    gy  -            dx  5    3  x2  —  10  axy 

*Vl-z2'  '    5  ax2- 21  j/2" 

t(fes-^i?  +  V1-^  2x(l-y2)*-3y(l-y2) 
x2Vl-x4 


(to.  ^     _y2 


2y(l -y2)^  +  3x 


3     ,      5r6x8  +  35x2-12x-561 
3|_x*(x2-4)?(x+7)^  J 

4.    dy=V^V^r^^lg<fa.  8.    -5. 

2Va2-x2  y 

9.  Vx=_J2iL_;  ,„  =        120        .6V2:  6V& 
VlOO+y2  V100  +  y* 

10.  4  ft./sec. ;  8  ft./sec.  Ctenjp  ft  /gec 

11.  fVIT;  6f.  ra2         '' 

12.  17*;  IVU05.  lg     Vl6Tft./sec.2. 

13.  14  V3;  14.  50 

2  20.   —  2.24  lb./sq.  in.  per  sec. 

15.  Oy  =    c  k    ,  where  A:  =  Z)#.  21.  0.1  B0(a  +  2  &r). 

3«V  «o       flofr-2/rQ 

16.  y  +  ex2  =  0.  *    (1  -  eT1  +  r/12)2 

23.  (a)   ±?\/0;  (6)  0.899  a. 

4 

(c)  y  - 1 V2V3  -  3  =-(3  +  2  VS)  VWs  -Six-  ?). 

4 

24.  — in. /sec. 

9tt      ' 


ANSWERS  329 


Art.  56.     Pages  86,  87. 
1 .    —  a  sin  ax.  «/*  sin  6 


2.  3  tan2  x  sec2  x.  (1  -  cos  0)2 

3.  2  [cos  2  x  +  sin  2  x].  21.  -  w(^sinw«  +  5  cos  (at). 

4.  2  sec2  x  tan  x.  22  rsin^r1  + rcosd 

5.  6  sin  3  x  cos  3  x.  L        VZ2-  r2sin20- 

6.  6  x  cos2  (a2  -  x2)  sin  (a2  -  x2) .         23        sin  0 

7.  x  cos  x  +  sin  x.  *  ~~  cos  *P 

24.  a<p  cos  0  ;  a0  sin  <p ;  tan  0. 

25.  (a)  ^^;  cos0. 
w    cos  d 


8. 

?fxsec2-+4tan-V 
2^           2               2/ 

9. 

x        sin  Vx2      a2. 
Vx2  -  a2 

10. 

—  tan2  x. 

11. 

x2  esc  2  x  (3  —  2  x  cot  2  x) . 

12. 

x  sin  x.                 14.    2  sin2  x. 

13. 

x2cosx.                15.   3cos3x. 

16. 

2a(2sin0—  sin3  0) 

cos2^ 

17. 

a  sin  2  d 

Vcos  2  0 

18. 

l-w 

a  (sin  n0)  n  cos  nd. 

,,s.    a(\  —  cos  0)2    „  .    ., 

(o)  — ^ '— ;  a  sin  0. 

w  sin0 

(c)  a  sec2  -  cot-;  asec2-tan-- 
w  2        2'  2        2 

27.  tan  x  ;  a2  sin  x  cos  x. 

28.  90°. 

29.  arc  tan  (2  V2). 

30.  -  4-  nw. 
4 

31.  0.00027;  -0.00011. 

ee  32.    £  \/2a6. 

19.    a  sec3  -  tan  -  • 

3         3  33.    —  kv0  sin  Jet. 

Art.  60.     Page  92. 
1. - 7.    - 8.   0.        9.    1. 


Va2  -  x2                                               P  v'p«  -  a2 
V&  10. 


A;2  sec2  x  tan  x 


'    &+x2  (2-&2tan2x)Vl-fc2tan2x 

„  2  a2  11.   2  x  fare  cos  x2 ^     \  . 

3     x./5*37S"  V  VT^j 


*  v  x*  -  a4 


12. 


4 ? w  (ra2  +  x2)  " 

V*nr^Vi-a2  +  x2  13   _Lv-rr*. 


X 


x- 


5.   arc  tan  x  -( i     . 

1+x2  14.    iv/x2-a2. 

x 
a  a  x2 

b- .  15  16.  arccosx. 

/>  Vp2  -  a2  Va2  -  x2 


330  ANSWERS 

Art.  65.     Pages  97,  98. 

1.         2*-3      .  12  1 


x2-3x+5  Vx2  -  a2 

2 2m  1 

(*-*,»)  loga'  13-    j^T*' 

3     xa^-^loga  14.   arc  tan  x.  15.   xeax. 

Vx2  —  a2  16  x 


4.  3  x2e*3  +  e*.  a  +  &x2 

5.  1+logx.  '  17     l-2x  +  2x2-3x3 
6          1  Vl  +  x2 

xlogx  lg     x(x3+9x2-24) 

7.   e*-e~*.  '     Vx^4(x+3)3* 


4 


(ex  —  e~x) 


X\t 


19.        *'  +  * 


9.   e8inxcosx.  x  (1  -x2) 


in     sec2x  20. • 

10-    ^nV  (l+*)vT=^ 

n  2  21.    Vl-C^e^*-^-0^). 

Vx^TT*  22.    e-M(B-Bkt-Ak). 

23.  e-**  [(ma  —  &&)  cos  m£  —  (mb  +  a&)  sin  mi]. 

24.  £?:  ae«»;  5f!±^L±i;  e.«>  V^+T. 

a  a 

25.  45°  ;  arc  tan  -  ■ 

e2 

ea  —  e  » 
Miscellaneous  Exercises.     Pages  98-100. 

4.  (a)    j  +  wtt;   ±J;   §tt  +  wtt.  (6)    !£;  arctan(±2);   ?  +  ?• 

4  4      4  2  4        2 

(c)   ~  +  wtt  ;  arc  tan  (±  V2)  ;  y  +  n?r.     (d)  x  =  1  ;    -  ;  nowhere. 
4  4  4 

5.  (a)  0.80902.  (6)    -0.58779.  (c)  1.52786. 

6.  sin*     ;  0.5773;  r.  10.    1 

l-cos0  x4(l+x2). 

V«2  -  x2  pX  _  P-x 

11.  <r-e     . 


Va2- 

X2 

X 
X2 

V2ax 

+  x2 

1  1 

12        ax  sec2  ax  log  a 

9.    Va2  -  x2. 


ANSWERS 

13.    (a) 

a(l  +  coB0)«       Bin, 
sin  6 

(»)  -';  ?■ 

a       2 

00 

-  p  tan  0  ;  a2  sin  2  0. 
2P 

0                0 
(d)  />cot-;  ptan -• 
z              2 

14.    (a) 

1  +  cos  0             ,,, 
sin0 

2 
a 

(e) 

2                 V'         2 

331 


16.   ?»cos-;  msin- 
2  2 


17.  (a)  -j^JL  +  i^V      (6)    T(^nfcy      (c)p(ba<>  log  a-  cplogfl). 

18.  —  ae-*{2ir&  sin  2  «■(&«  +  c)  +  \  cos  2  *•£&*  +  c)}. 


Art.  68.    Page  103. 
6.    f  x7  -  x4  +  x2  +  7  x  +  C. 

7-  -i+**-f+X  +  a  9.^-2*1  +  a 

10.    M^  -  12  x8  +  ^^  -  36  x6  +  §i^f  +  a 
9  7  5 

(ax +  6)3  (2^-5)*      c 

3  a  32 


12 


(x3  +  4)«  |   c 


14.    (3^-^+C, 
30 


3  15.    2Vx3-5x  +  7  +  O. 


Art  70.     Page  108. 

1.  $  (x  +  ay  +  C.  13.    v^T^2  +  a 

2.  i{x*-a*y  +  c.  14   iarcsin^  +  a 

3.    log  (3  x2  +  a2)  +  0.  c 

4#  ex2  _j.  (>#  15.   arc  sec  ax  +  C. 

5.  --L+c,  16.     «w2'3  +a 

sin  6  m2  i0g  a 

6.  -icos3  0+O.  ,  j 

7.  Karctanx)2+C.  17-    i  arc  tan -^- +  C. 

8.  -  -  VI  -  m2x2  +  G.  .    x-± 

m  18.    arc  sin  — \-  C. 

9.  i  tan2  0  +  C. 

11.  A(«x  +  &; 
7  rt 

12.  ^log(2x8-5)+C.  12     "11-x 


9  v  19.    log(x  +  3+Vx2  +  6x  +  10). 

11.    ^-(ax  +  b)*+C.       •  bK 

1(1  20.    JL]ogJL±£,tf-l<c<lL 


332 


ANSWERS 
Art.  71.     Page  110. 


1.    75O  +  bx~  a  log  (a  +  to)]. 


4.    f(4  +  x)Vx-2. 


9 

VX2  -  «2 

? 

a2x 

X 

a2  Va2  -  *a 

5.    log 


V2  x  +  1  -  1 


V2  a;  +  1  +  1 


7.   arc  cos 


Miscellaneous  Exercises. 


-^-  *«-2+  a 

a  —  2 

2V3x-x2  +  a 

-2     +a 


10. 

11. 
12. 

13. 

26. 
27. 

28. 


3y/x*-a8 
logvx2-6x  +  1  +  a 


Vx2  -  ti  x  +  1  +  a 
|  e*2  +  a 
- 1  cos3  e+c. 
- \  esc2  0+  a 

2  arc  sin  -  —  3V4  —  x2. 


a 


aVax  +  6 
&log(3x2-5)  +  C. 


log  vx4  -  5  x2  +  2  x  -  7  +  C. 

1  arc  tan  £±J!  +  c. 

2  2 


14. 


Pages  111,  112. 

1  xs  -  1  a;2  +  x-  log  (*  +1)  +  C. 


15. 

log  (x3  +  x)  - 1  log  (x2  +  1)  +  C. 

16. 

-  -  log(a  +  &  cos  x)  +  a 
0 

17. 

—  e008*  4-  (7. 

18. 

arc  tan  x  -  log  Vl  +  x2  +  C. 

19. 

log^  +  l+C. 

X+  1 

20.  arc  sin  2a;  +  3.  +  (7. 

V57 

21.  log  [x+  8  +  Vx2  +  6  x  +  1]  4-  C. 

22.  J-logL=LV?+a 

2V6        z+V6 

23.  — i--iogag£=J+a 


24. 
25. 

c  log[w  +  wi  +  V(w  +  ro)2-  w2]  +  C. 

33. 


4V3        V30  +  2 
arc  sin  (log  x)  +  C. 
log  [s  +  a  +  vV2  +  2  «s]  +  C. 


i  arc  sin  —  +  C 


y  =  log\/x2+  1  +  2. 


29.    log 


arc  sin  x  +  V 1  —  x2  +  G. 
1 


34.    y 


=  ^(x2-l)  +  l-I 
2  X 


+  a 


l  +  r 

2(Vx_-L\  +  C. 
v  vx/ 


35. 


s  =  ^  sin  to. 


ANSWERS  333 

Art.  73.     Page  114. 

1.  y  =  3x+C;  y  =  lx*+C;  y  =  ^mx'2  +  C;  y=l+C;  y  =  lax*-bx+C; 

x 

y=  -  -i-  +  G. 

y  2  ax* 

2.  y  =x2+  C.  5.   6  =  kp  +  C.  8.  y  =  eax  +  C. 

3.  y  =  bx  +  C.  6.  m6  =  P  +  k.  9.  log  y  =  -  +  (7. 

a 

4.  y3=mx+C.  7.  y  =  f  x2  +  5x- 13.  10.  p=ea*. 

Art.  76.     Page  118. 

1.  (a)  v  =  v0  +  mt  —  ^  nts ;  (b)  v  =  v0  —  mk  sin  kt ; 

s  =  So+Uo*  + 1  mt2  —  ^  nt*.  s  =s0  +  v0t  +  m  cos  to. 

(c)  v  =  ^fe*  +  e-*0:s  =  —  4-   8  sec. ;  15.27  rev. 

2  &2' 

2.  a  =  10-2«i;  f  =  6tf-lfc«.  7.  y  =  -  ^       8.  85.12  ft./sec. 


3.    10  sec;  166§  ft.;  -10  ft./sec 


Miscellaneous  Exercises.     Pages  124,  125. 

1-    ^  arc  tan   *.  +  C.  4"   arc  ten  e*+  C' 

6                  V3  5.    ±  arc  sec  I* +  C. 

2.  p  +  .  +  jwg+a  V7                V7 

. „  6.    i  arc  sec  ^=^L  +  (7. 

3.  fvV  +  l(e*-2)  +  C.  6                   ft 

?•    (a)  y  =  fx2  +  2z+C.  (6)  ?/ =  f  x*+ Cx+ C". 

(c)  y  =  sin  mx  +  C.  -. 

8.  t,  =  4s-*rf-6.  W  f  =  -#-+<7'. 

9.  Q  =  ar +  %bT2  +  $CTs.  13.   p  =  ce«. 

K                     6  w      2          3        12 

W       16    '"  24  '  ^  ;       16        "    6         48  ' 

Art  82.     Page  129. 

1.    6x.  4    _       x 


3 

2.  x»  ri  +  (log  w  +  i)2i.  c1-*2) 

U  J  .  a2 

o.    — 

3.   eax[(a2— 1)  cosx  — 2  asinx].  (cP—o?)^ 


334 

6 

2(a2-x2) 

7. 
8. 

(a2  +  x2)2 
e*(x  +  3). 
2 

ANSWERS 


12.    (-I)'"' 

xn+l 

13<    (-l^-Kn-l)!, 
xn 


(x-3)3 

9.    -(xcosx  +  3sinx).  14.    (a)   _£?!.       (6)   _  _&_ 

11.   a*(log  a)».  */3  fV 

15     12(x7y  +  7  x6y2  +  23  x5y3  +  38  x*y*  +  23  a%5  +  7  a?V  +  xy7) 
(xs  +  6  x2y  +  3  xy'2)s 
(a2- l)Q/2-2ax?/ +  x2).  1ft       2  +  cos0 


16.  v —  A  J  vy —  *  m-*^  -r  j^y .  18 

(y  —  ax)3  (1  —  cos  <f>)2 

17.  i.  19.    2  cos2  0  cot  0. 
x2 

Art.  83.     Pages  131,  132. 

1.    2>-4(sin  ax)  =  -sin  ax  +  ^x3  +  ^x2  +  c3x  +  c4- 
a4  6  2 

2*    2>"4  ==^*a  "S**  +  \ClxS  +  lC2x2  +  CsX  +  C4* 

3.     I)"*  =  —  |  log  X  +  £  CiX3  +  |  C2X3  +  c3x  +  c4. 

4    y  =  ^8     tn^-Bw^ 

6  6  to 

5.    #/y  =  \  Mx*  +  I  i?x3  -  &  wx4  +  (7ix  +  C2. 
C2  =  0  ;  Cx  =  -\Ml-\  Bl2  +  &  wl\ 

Art  84.     Page  136. 

1.  Max.  for  x  =  0  ;  min.  for  x  =  6$.  11.  Min.  for  x  =1 7r,  1 7r,  etc. 

2.  Max.  for  x=0;  min.  for  x=  ±  V8.  12.  Min.  for  x  =  2. 

3.  Max.   forx  =  -iV3;    min.    for  u  Min.  for  ,,  =  _  2J> . 

x  ss  £v3.  3  a 


4. 
5. 
6. 
7. 
8. 
9. 

Max.  for  x  =  0  ; 
Min.  for  x  =  2. 
Min.  for  x  =  1.  . 
Min.  for  x  =  v2. 
Min.  for  x  =  0. 
Max.  for  x  =  1. 

min. 

for  x 

f.      14.   Min.  for  x  as  —  f. 

15.  Min.  for  x  =  —  §. 

16.  Max.  for  x  =  e. 

17.  Min.  for  x  =  3  and  for  x  =  —  2. 
Max.  for  x  =  —  f . 

18.  Max.  for  x  =  0. 

10. 

Max.  f  or  x  =  -  • 
4 

Art. 

85 

Pages  138,  139. 

1. 
3. 

2  and   *. 

2            2 

-  p  +  Vl  +  ac2. 

5. 

2.   45°  ;  max.  range  =  ^Sl  . 
Q 
8  in.                                                 7.    V15. 

4. 

a 
6* 

6. 

Length  =  diameter.                       8.   4  V5l 

ANSWERS  385 

10.  Altitude  =  2  x  diam.  of  sphere.  lg     e 

12.  Base  =  altitude  =  rV2.  '    2* 

13.  a  a/2,  6V2.  19.    ff. 

14.  Base  =  a\%  altitude  =  §  a.  „ft  ,  ,  _  14  V3 

15.  Altitude  =  |x  altitude  of  cone.  breadth  _  ___ . 
Radius  of   base  =  f  x  radius    of  14  ^g 

base  of  cone.  depth  =  — - — 

16.  Altitude  =  f  a ;    radius  of    base  ^     „       ,  ,       n     _  - 

_Ia^§  21.    Breadth  =  6;  depth  =  6V3. 


17.    Radius  of  base  =  ^-£-v/— ; 

2_V2  7r' 

altitude  =  yj—  ■ 

Miscellaneous  Exercises.     Pages  140,  141. 

1.  sin  *(2  -  **)  +  4  *  cos  x.  g     (a)  4  e,(sin  x  _  cog  %)  .  (ft) 

2.  26*cos(*+j).  ^gffl^ 

|  '    *  24^7 

a3 

3.  — — —  •  10.    y  =  ax*2  +  cix  +  C2  ; 


S&M 


y  =  axJ  +  cix  ; 


4         3a4x  y  =  ax2  +  5  ax  -f  c2. 

(x2  -  a2)  * 

11.  (a)  I)-»  =  -.|loga;  +  iaB*  +  ^  +  lc1»*  +  Cjaj  +  cg. 

A  Z*x  O  — 

(6)  Z>-3  =  A-  (e«*  +  e-*8)  +  icix2  +  c2x  +  c3. 
a3  2 

re)  7>"3  =  i  cos(kt  +  e)  +  ;Ux2  +  c2x  +  c3. 

A/  2 

12.  (a)  a2cie-*' + /32c2e-^. 

(6)  ae-«[a(ci  +  c2£)  -  2  c2]. 

13.  e-«'[(a2  +  /S2) (ci  cos  pt  4-  c2  sin  /3<)  +  2  «0(ci  sin /S<  -  C2  cos  /»)]. 


15.  J_  ^      -^  i-Vl-c^. 

*  VI  -  c2 

16.  (a)  Max.  for  x  =  3  ;  max.  value  =  2. 

Min.  for  x  =  —  1  ;  min.  value  =  |. 
(6)  Max.  for  6  =  £  v  ;  max.  value  =  £  V3. 
Min.   for  0  =  1 7r  ;  min.  value  =  —  f  V3. 

17  1  18     a6c  +  2fgh  -  af2  -  bg2  -  ch* 

a(l-cos0)2  (hx+by+f)* 

19.    r»Aco*e  +  jV±*™y-~\. 
L  (7J8  -  r2  sin2  0)£J 


336  ANSWERS 

Art.  86.     Page  143. 

1.  Concave  up  if  x  >  §  ;  concave  down  if  x  <  §. 

2.  Concave  up  if  —  -  +  2  rnr  <  x  <  |  -f-  2  nr  ; 

2  2 

concave  down  if  -  +  2  nr  <  x  <(2  w  +  l)ir. 

A 

3.  Concave  up  if  x  <  —  1 ;  concave  down  if  x  >  —  1. 

4.  Concave  down  at  every  point. 

5.  Concave  up  at  every  point. 

8.  (a)  Concave  up  at  every  point,     (b)  Concave  down  at  every  point. 

9.  Concave  up  at  every  point. 


Art.  87. 

Pages  144,  145. 

1. 

2. 
3. 

4. 

x  =  h 

No  point  of  inflexion. 
No  point  of  inflexion. 
Points  of  inflexion  for  x  = 

7.  x  =  0. 

8.  x  =  0. 

9.  %  =  \y/2. 

flTT.                            10.      X  =  ±  1. 

5. 

Points  of  inflexion  for  x  = 

»Z.               11.   x  =  ±J±- 

2                                    V'Sb 

6. 

x  =  log  2. 

13.   ?/  =  ix3  +  x  +  3. 

Art.  88. 

Pages  148,  149. 

1. 
2. 
3. 

4. 

5. 
6. 
7. 

8. 

y  =  x-\b. 

x  =  a,  y  =  ±(x  +  a). 

x  =  2a. 

y  =  ±^x. 
a 

x  =  2. 

3x  +  3y  =  2. 

x  =  ±c;  y as ±  c. 
x  =  l. 

9.   x  =  a;  y  =  x  +  ^;  y  =  -x- 

10.  y  =  0. 

11.  y  =  x  +  2. 

12.  x  =  0  ;  y  =  0. 

13.  2/  =  x+f 

14.  x  =  0  ;  y  =  0  ;  x  +  y  =  0. 

15.  x=-l. 

16.  No  asymptotes. 

~2 

Art.  89.     Pages  151,  152. 

1.  Double  point  and  cusp  at  (0,  0)  ;  tangents  y1  =  0,  y  =x. 

2.  Cusp  at  (0,  0)  ;  tangents  y2  =  0. 

3.  Tacnode  at  (0,  0);  tangents  ?/2  =  0. 

4.  Tacnode  at  (0,  0);  tangents  y'2  =  0. 

5.  Double  point  at  origin  ;  tangents  y  =  ±  x. 

6.  Double  point  at  origin  ;  tangents  y  =  0,  x  =  0. 

11.  Cusp  at  (a,  b) ;  y  =  a  the  common  tangent. 

12.  Double  point  at  (0,  0) ;  tangents  y  =  ±  Vax. 


ANSWERS  337 

Art  92.     Page  160. 

1  (y2  +  q2)l  g     (4y*  +  aQ*  Vs 

a2  2  */4  —  4  x% 

2  (9a2y4  +  4x2)*  ^    aVa;(8a  -  3^ 


3  ay (6  ay*  -  8  x*)  '        3(2  a -z) 

+  W)1  f1     (x+j 

a464  "'    ~       i 


fCar+a^  +  l]1  12.    -15V3;    (20,   -  10  V2). 


13 

Qgff;    (21*,  16*). 

14. 

-  1  ;  (0,  0). 

15. 

1 
2  ak' 

1R 

2a 

4. 

x  +  a 

5.  sec  x. 

6.  3(a«y)i 

T.  «?• 

a 

a4&4  [l  +  (2ax+6)2]* 

17.   z  =  |  log  £  =-  0.1534. 

Art.  93.     Page  162. 

I.  J_(a2sin2^  +  &2cos2^)2.  2.  3  a  sin  0  cos  0. 
ab 

3.  y  +  rf)1.  5.    ?asin2|.  7.    fa. 

P2  +  2a2  4  3  *a 

4.  jl.  6.  pvrr^.  8-  i- 

V2  ° 

Art.  96.     Pages  166,  167. 

1.   The  origin.     2.  4(ro  -  2/>)3  =  27  pn*.    3.   faro)*  +  (&n)$  =(a2  -  6*)$' 

4.  m^q2-fi2cos«g,  n=-^=A2sin80. 

a  o 

5.  m  =  a  cos  0,  w  =  a  sin  0,  a  circle. 

Miscellaneous  Exercises.     Pages  167,  168. 

'•-^(f'-f)-  »•  ■-»'  (0-0)- 

II.  Double  point  at  (a,  0)  ;  tangents  y=  ±(x-d). 
12.    Conjugate  point  at  (0,  0). 


338  ANSWERS 

13.  Double  point  at  (0,  0)  ;  tangents  tj  =  0,y  =  *x. 

14.  (a)  |f  =  *  +  |;  (6)  y=±^x.        15.   -8asinf0. 

a  2 

16.  ±x  =  a\oga+^-*-vanrj2. 
y 

18.    (a)  P  =  a  sec  0 ;  (6)  the  initial  line  6  =  0. 

(c)  four  asymptotes,  p  =  |  sec  fc ±  »\ ,  />  =  |  sec  (£*  ±  ^  . 
20.   4096  asm  +  1152  a2n*  +  27  n4  =  0. 

Art.  97.     Page  170. 

9.   logV3.  13.   2. 

17    - 
10.    V2-1.  14'    ^arcsin  -•  '4* 


I*,  i-                                15.   t(a*-l)8  18.    10^10-1 

12.    1.                                     16.    2.  64 

Art.  99.     Pages  172,  173. 

1.   |  arc  tan  4.                    4.   K*4-!)-  7.    ilog3  2. 


5.   arc  tan  e  —  *  •  8.       - 

«2V2 


4 
3-    f  6.    1. 

Art.  103.     Page  182. 
X'fa'  2'i  3-    -  *-|-  8.|(1+^T5). 

Miscellaneous  Exercises.     Pages  182,  183. 

1.    f\/2~^T3.  9    2x/2 

?  la8" 

3.    &,  2 

5.  (a)  i(cos  a  +  cos  /8)  ;  (b)  §  a.       IX.   ao. 

6.  e*-e. 

Art.  105.     Pages  187,  188. 
L    78-  „     4  7r3q2 

2.  106f.  °*    ~~ 3 

3.  10.  9.    «2 

4-    32.  10.    4  7r2a2. 


a(ea  —  ea). 


11.  fa&. 

12.  a2. 

,2 


7.    J^(6^_^?);01og».        14.   t 
m-\\  J'         °a  n 


ANSWERS  339 

Art.  106.     Page  190. 

1.    (a)  1 7rw2m  ;  (ft)  }  irm2n.  2.   27  *-. 

3.    144  r.  5.   |-7ra2(ec  +  2c-e-c).  7.   f  *-a3(31og2  -  2). 

13  g 

Art.  107.    Page  192. 
1.    108  7T.         2.   fTra&c.         5.   a2/*(>-A).         6.   698f  cu.  in.         7.  240. 

Art.  108.     Page  194. 

1.  6a.  ,  ■,       zh\ 

27  LV       4aj  J'  4.    2*-a.  5.    §*%. 

Art.  109.    Page  196.  

1.2. a.  3.    Vl  +  «2- 

2.  VlS/.y.tX.    VS2/^...?\.  4.   a.    ' 


^(.r-i), l^(— .?). 


Art.  110.    Page  198. 
1.    §2_Z(2V2-1);  ^(5V5-2V2).         3.    |j[(87)*- 1]. 


27 


2.    7rwVw2  +  w2.  4.    EiL(e2_e-2  +  4). 

4 


1.    |V5. 


Art.  111.    Page  200. 

4     7rW'2, 
6 


2.  -5f.  5.  pi-^- log^2. 

z2  —  X\       %i 

3.  (a)^;  (6)  0.  6.    2jr. 

7T  „ 


Art.  113.  Page  204. 

1.  4303$in.-lb.  5.   83,219  ft.-lb. 

2     Mh4-mh2. 

2.  Mh+—-  6     656g4  ft_lb< 

4.    TF=-(s28-si3). 

3W  y  7.    3,273,000  in.-lb. 


340  ANSWERS 

Miscellaneous  Exercises.     Pages  204-206. 

1.    (a)  372;  (6)  48;  (c)  e-1. 


kS. 

8 

3. 

log  2. 

4. 

a2 
6' 

5. 

00  If*;  (P)  1*. 

6. 

128  cu.  in. 

8. 

fa3. 

9. 

a2. 

10. 
11. 

141.  (73)* -27 

16 
f  mn2. 

12. 

2io 

13. 


2p0 


14.  Clog^-6-0^2-^. 

15.  *(*w+1  -  *iw+1) 
w  +  1 

16.  56,530  ft.-lb. 

Art.  114.     Page  209. 

1.  e*(x2  -  2  x  +  2)  +  C.  5.   f(*  -  sin  6  cos  0)  +  C. 

2.  a;m+1  i0g  x  _   s  OT+1     4.  c.  ft    qea3;  sin  &a;  —  beax  cos  ^  4.  n 
m  + 1               (m  +  l)2  a2  +  62 


3.    0  arc  sin  0  +  Vl  —  02  +  C.  7.    x  log  x  -  x  +  C. 


4    0  arc  cot  0  +  log  V 1  +  0s  +  C.  8.  sin  0  (log  sin  6  -  1)  +  C. 

9.  gMetm»-g+?°£ig±31+g, 

3  6  6 

10.  "VdnS  +  ooiSUc. 

2     V       a  «/ 

11.  cos  x  +  x  sin  x  +  C. 

12.  *  ["tan  0  sec  0  +  log  tan  (|  +  ^  1  +  C. 

13.  f[V2  +  log(l+  V2)]-  14.  ?~ 
15.   |[(logx)2-|logx  +  |]+a         16.   |. 

Art.  115.    Page  211. 

2.    log  V(x  +  5)5(x  -  3)  +  C. 

«    logx*(x-4)*      ^  ,/«TXi 


ANSWERS 


341 


i  a2  -  x  +  |  log  x  +  H  log  (x  -  5)  -  §  log  (a;  +  1)  +  0. 

log  j-ftii^JQ  j  +  (7.  8.  log  [*2(a  -  2)*(*  +  2)^9]  +  (7. 

Art.  115.     Page  212. 

\og*^*  +  -L-+c.  5.  iogtri-.i+a 

*  S-2  y  J 


'-vr-^r-^j^..  ,log 

9~4x    +  0.  7.   log 

2(X  -  2)2 


+ 


(a;-!)8      1 

x*  x  '  x  -  1 

x-1 


+  C. 


x  +  1      x-1 


C. 


log(x  +  3)+-4_  +  a 
x  +  3 

logx-^4-  C. 


8.   lie**-1 


4       x  +  1      2(x+l) 
10.  mlog(TO  +  x)4-  2m* 


+  0. 


+  x     2  («»  +  x)2 


+  C. 


Art.  115.     Page  213. 

log  (s-l)3(a  +  l)2  +  arc  4  tan  j.  +  0 
(re2  + 1)» 


I  log  * — 1  _  I  arc  tana;  +  C. 

4        x  +  1      2 


4.  I  log  ^±1  + a 

4        x2  +  3 


3. 
6. 
8. 
9. 

11. 

12. 


5arctan* ■ +  I+0.  5.   1  log  *±I+C. 

2  2(l+x2)      x^  6     8x2  +  4^ 

7.   |log^±|  +  |log(y  +  l)  +  -4arctan^+C. 


8     &  y  -  1      8  '  V3 

■»f«    4-gawtanfUa 

Vx2  +  62      &  &J 


V3 


_lrI  +  J_arctan-*-]+C.        10.  log  ^S  +  arc^nx  +  o. 
log  (l  +  *  +  *2)f L__  JL  arc  tan  2-^±l  +  C. 


(1+X) 


1+*         V3 


log 


VI +2  a? 


+  0. 


V3 
13.  log  v/x2TT  + 


2(x2  +  1) 


+  G 


Art.  116.     Pages  217,  218. 

3[f-^  +  |-^  +  ^-log(,Ul)J+C. 

r4x*      4_aj      64xj      128s*      1024^      4096 1     (3ai_4)-]  +  a 
L   5  3  27  27  81  243      BV  J  J 


342  ANSWERS 


3     2  r(mx  +  b)%     2  6  (mx  +  b)$  .  b2(mx  +  b)?l      G 
mA.        7  /5  3  _T 


4.  2arctan-Nf55+6\  6'  log  [x  +  |  +  V^+5x-3]  +  a 

7.  log  [x  -  |  +  Vx2  -  7  x  -f  4]  +  C. 

5.  6  arc  tan  J^=-^  +  0.  8.  J^  arc  tan  J3(^-2)  +  a 

'5  —  x  f8  ^2(3  —  x) 

9,  2j(^-l)l+§(^)l+(,.l)l+(^.1)l|  +  0. 
10.  3  Vx2^  -  log  [x  +  Vx2  -  3]  +  C. 


11.  log[x  +  l-fVx2  +  2x  +  5]+C. 

3 

12  (i-x2)^_(1  _ x2)|  +  0  15  log [a.  + 1  +  Vx^T^]+  c. 

o 

|8.     *2  +  2    +C.  16.   -V^g-K?, 

Vx2  +  1  a2x 


14.   _  2V2x-x2  _  arc yersx  +  c        17  x         _  arcgin  x  +  c 

x  Vrt2  —  x2  <? 


18. +  log(g+Vg2-qa)  +  a 

Vx2  —  a2 

19.  |  Vx2  -  a2  +  —  log  (x  +  Vx2  -  a2)  +  C. 

2  2 

3«4 


20.    ^(2x2  -  5a2)  Vx2  -  a2  +  ^ii- log (x  +  Vx2  -  «2)  +  C. 
8  8 

21.  f  arc  sin  x  -  2x*  +  Zx  VT^tf  +  C.  24.   ^-2. 

8  4 


22.  f(a£-2)vT-H?+  C. 

23.  ^x^ri_(x*-l)*      a  25     11V16 

x  3x8  64 


Art  117.  Page  221. 

1."  I  sec6  x  +  C.  7.  \  tan2  x  —  log  sec  x  +  C. 

2.  —  cot  x  —  £  cot3  x  +  C.  8.  -J  tan6  x  +  C. 

3.  —  £  cot5  x—  \  cot8  x  +  C.  9.   —  sin  x  —  esc  x  +  C. 

4.  -$cot8x+0.  10.  ftan^x+O. 

4  JQ 

5.  -  $cos3x  +  I  cos5x  +  C  11.  f  sin^x  —  T%sin  *  x  +  C. 

6.  —  )  cot2  x  +  log  tan  x]+  C.  12.  tan  x  —  cot  x  +  C. 


ANSWERS  348 

Miscellaneous  Examples.    Pages  228-230. 

1. — (-  +  2x\  +  C.  3.    &x™-\x*+C. 

Vl  4-x'2\x  I  -  3  -  1 

4x_3  4.    t(V*+o)i-4a(Vx+a)*+C. 

2'    Vi"0  Si"  ~VU  +  °-  6.   (*-l)vtt-g+T  +  ft 

6 

6.  (0  +  c)  cos  e  +  sin  e  log  sec  (0  +  e)  +  C. 

7.  (a)   -  (5  a2  -  2  x2)  Va2  -  x2  +  ^  arc  sin  -  +  C. 

8  8  05 

(6)   -  (2  x2  -  a2)Va2-x2  +  ^  arc  sin  -  +  C. 
8  8  ct 

9.    (a)  itan4z-i  tan2  x  +  log  sec  x  +  C.         (6)  |tan30-  tan  0  +  0  +  C. 

10.  log ^/fo-4)10  +  C.  11.    i  x3 arc  cos  a; - |  Vl^x2  + 1(1  -  x2)  *  +  C. 

'     x  —  1 

12.    |earctanx    1  +  x    +  (?.  13.    i  e^l  +  2  sin  x  cos  x  +  2  cos2  x)  +  C. 

Vl  +  x2 

H.    ![V^-a2arccosf]+a     22.    -^bgj^+C. 

21.    _±_  log  ^&±J&*  +  d       23'    («  +  **>*  (£  -  M2)  +  " 

2V2  Vl  +  x2-V2x 

25.  *■«&.  27.    3  r#.  29.    8(6- a)*. 

26.  |*<l&.  28.    10  log  10 -9.  15Vwi 

30.  (a)    V^TT  +  2  -  V  2  +  log     *  +  ^     . 

1  +  Ve*  +  1 

(6)   V37  -  V2  +  log  6^  +  ^)  •  (c)  2V3. 

1  +V37 

31.  |[7rv/TT^2  +  l0g(7r  +  Vr+7)]. 

32.  2arv^-2-V31og  v^+v^n 

L  2V2  +  V6J 

33.  f  Tra.  36.    5  n-2a3.  39.    V  ™8- 

34.  12  a.  37.   8  7r3a3.  40.    fa2A. 

35.  |9ra62.  38.    \  tt2. 

41.  2  TrftTd  +        a2    -  arc  cos  -1 ,  2  imfa  +      '  62       arc  cos  -1- 

L         Va2-//2  «J  L        V62  -  a2  bJ 

42.  ^a2.     43.    Sf.       44.    if.       45.    BT\ogP±- C(n  ~  l\Pln-p.2»). 

5  4  tt  pa  n 

46.  „=*p^£zi£;  «=J5Trv^r^?- &faic cob fitzJi-jU 

Vo  a  \2&L  2  V  s0  2/ J 


344  ANSWERS 

Art.  123.     Pages  235-236. 

1.  2xy5;  5x2?/4. 

2.  cos  x  cos  y  ;  —  sin  x  sin  y. 

3.  yex  +  ey  ;  ex  +  ze2'- 

4.  4  a3—  2  axy  +  6y2 ; 

—  ax'1  +  2  &x?/  +  4  ?/3. 

5.  yxlog?/ ;  xy*-1. 

6. * :   * 


Vy2  —  x'2     y  y/y2  —  x2 

7.  y  cos  x  +  sin  y  ;  sin  x  +  £  cos  ^. 

8.  cos(x  +  z/)  ;  cos  (x  +  y). 

10.   ex  log  yz;   -;    -• 

y      z 

19.  m(m  —  \)xm~2yn  ;  mnxm-lyn~l  ; 

20.  6  x ;  a  ;  a ;  -  6  y. 

22.    0;  -3?/ 


11. 

cos  x  cos  ?/  —  sin  x  sin  0  ; 

cos  y  cos  0  —  sin  x  sin  y  ; 

cos  x  cos  2  —  sin  y  sin  0. 

12. 

1.1             1 

x  +  y'   x  +  y'       z 

13. 

du  _2x*-if-  z* 
dx             x2yz 

14. 

2wrh.   irr2 
3     '     3 

18. 

0;   I;   1;    -*. 

v    y      y2 

nxm 

-iyn-i .  n(ji—  l)xmyn-2. 

21. 

ex+y  .    ex+y  .    ex+y  .    e*+y# 

-3 

y2;  -6xy. 

Art.  125.    Page  241. 

1.  2{c  +  cos xe2 sin x) .  6.   Increases  3|  sq.  units  per  sec. 

„    ea(x  —  1)  7       679     units 

x2  +  e2x  '    15VTI   sec 


3-       ,  8.    -0.284 


cu.  ft. 


Vx2  +  a2  ■   sec 

a      Lo  -2   •      x  q     C(l-w)     n-l"/C 

4.  e*(2  x cos x  —  cos x  — x2smx).  9-    — ^— -;   — - — \  —  • 

5.  ea*sinx(a2  +  1). 


i?v"  ita     >j0 


Art.  126.     Pages  243,  244. 

1.   exy2  dx  +  2  exy  dy.  2.  y^  log  y  dx  +  xyx~1dy. 

3.   cos  x  cos  ydx—  sin  x  sin  ?/  efy.  5.  axe*(\og  adx  +  dy) . 

.     2xdx     3  x2  cZy  6  ydx  —  xdy 

y*           t  x2  +  y2 

7.  3(x22/*  +  i  x" %y*)dx  +  K*V*  ~  4  x'^y)dy. 

8.  (ydx-xdy)(— i— +  — — V 

U2+2/2     xy/x2-y2> 

10.  £**  log  z(y  dx-\-x  dy)  +  xy  s^  -1  cte. 

11.  cos  x  cos  y  tan  z  dx  •*■  sin  x  sin  ?/  tan  z  dy  +  sin  x  cos  y  sec2  2  dz. 

12  zydx  +  zxdy-xydz  M    d       B(dT_Tdv\ 

Z2  +  X2?/2  '  \  V  V2    / 

13  2X^^+X2^+3X2^^.  15       dk=C(Tndp+npTn-ldn 

a*  —  ys  (a3  —  y3)2 

18.    Approx.,  0.00117;  actual,  0.00116.         20.    bAa  +  aAb  . 

aft 


ANSWERS  345 

Art.  127.     Page  245. 

x  6xy-y2  4  4  x(x2  +  y2)  +  y2  7    _£ra_ 

3y2-3x2  +  2xy'       '        4*/(x2  +  y2)-\-2y{x-  2  a) '  b  -  v' 

„    3x2siny+y3sinx  g    4  a;3  —  3  y2  8        p  sin  d 
3 y2 cos x— xs cosy        '   6y(x  —  y)  '    2cos^— 3p 

3.    -^--  6.    -tan*,.  9.    -^. 

wx  «2y 

Art  128.    Page  251. 

1.  Exact.     ex  sin  y.  6.   Exact.     \  ys  +  xy2  —  (x2  +  x)y. 

2.  Exact.    pvn.  8.   Exact,    xsiny  —  exy. 

5.   Exact,     ^-xy.  11.    («)   -4;  (&)   +4;  (c)  -8. 

Miscellaneous  Examples.    Pages  251-253. 

1.    2xy3z5;   3  x'ty225 ;   5x2i/¥. 

«  x2  y2  z2 


(a8  +  2/:}  +  z3) !      (xs  +  y8  +  z3)  *     (x3  +  y3  +  z8)^ 
3.    3  x2(?y  cos  z  ;  2  xM2*  cos  z  ;  -  x3e2*  sin  z. 


4. 


x  +  y 


^  +  y)-2  +  g?         (X  +  yy2  +  38  (^  +  ^2  +  ^2 

5            4  y4  -  3  x2y2  6  ^x                   ?         ax  +  hy  +  e 

2  xsy  -  10  xyz  +  15  y4  '  a2y                             by  +  hx  +  f 

„         Sx2y  9  a;  +  x3  4-  arc  tan  a; 

5y3-2x3  "  (l  +  x2)Vx2  +  (arctanx)2 

10.   3  xV  arc  sin  x  +  x8e*  arc  sin  x  +  -J**— .        11.   e~*(sm  x  ~  cos  x) 

-^/l  _  x2  e_2x  +  cos2  x 

12.    e«*sinx(a2  +  l).  13.   0;   -10;  72  y. 

14.  —  ez  cos  y  ;   —  e*  sin  y  ;  ex  sin  y. 

15.  x2?/  sin  xy  —  2  x  cos  xy  ;  xy2  sin  xy  —  2y  cos  xy ;  x8  sin  xy. 

19.   -*-;  ^ *!-.  21.    ?-E-;   ±    18 


on  «2  -  &2  +  & 

20-  :•  26.    V3;   20. 

cV4&2c2-(62  +  c2-a2)2 

27.  AF  =  7rr2AA+2  7rr/iAr;   M  +  2Ar. 

h         r 

28.  (a)  £*y(x2+y2)+0(x)+vKy).         (6)  i^ainy +  *■(*) +/(y). 
30.    (a)  x2y  -  $  y3.         (6)  (1  +  x2)  arc  tan  y  —  y. 

si.  («)  gr+gr,_^yi)^1  +  g^+,-o. 

(6)  aiogr+zsr-^Biogp-^p^i+lp^  +  Jo. 


346  ANSWERS 


Art.  129.     Pages  256,  257. 

1.    \&y*+F(x)+yKy).  g    yg5c2(4  +  c2)  . 


2.    —  e*cos  y  +  F(x)  +  <AQ/)- 


HI) 


3.    14,387f.  9.    fv^(***-*i*)fc 

10. 


4. 

7T 

8* 

5. 

*(»-t) 

fi 

a2 (8  -  tt) 

8 

7. 

fTTrt*. 

2 

11     ^* 
15  ' 

12.    J^ 
2  »  +  8 

Art.  130.     Page  259. 


3.  21i.  4    a2arcsinJi_^_&v/a2-&2. 

"  a2 

5.  |.(8*-8).                  7>   3.549.  9.    60.1+. 

„     1t>                                    8.    0.049.  10.    7ra6. 

Art.  131.     Page  261. 

2.  ^2.     3.   7r(n2-r.22).    4.   67m2.    %,  £L  7.   4.8584  a2;  11.1416  a2. 

Art.  132.     Page  264. 

4.  *iraba.               5.    ™*-c.               7.    i^8-  8.    faHan/3. 

3                             2                          86  ¥ 


Art.  133.     Page  267. 

3.   ?| 

2 

/Inrr  /'a/ 
4.      TT 


1.    fir*       2.    |L  3.    »£.         5.    |^(log8-2). 


-c^^-e-)-     fc^^^^jg 


Art.  134.     Pages  270-271. 

7ra3 

ttV2 


1.   Z  [C3  _  (c2  -  r2)2].  6.   I  Tra3  -fa8 

6 


2.  £  Tra&c.  7. 

2  •  16 

3.  Ja6c. 

Art.  135.  Page  273. 

1.    f  ka*;  7=i*?,  if  7  =  jty.  4.   |  Tr&a4 ;  |  fca. 

tt      „     „  .  5.1  kab2 ;  4  kb. 

3  6.    I  of  density  at  base. 
3.    \kl. 


ANSWERS  347 

Art.  137.     Pages  279,  280. 

£&$;•}  (»)(o,|i). 

2    (a)  /a[6V2  +  log(3-2V2)1;        4ar.2V2-11      \  /8a    §_a\ 

V  8[V2+log(V2+l)]       3[v'2  +  log(V2+l)]y  V  5  '    4^' 

3.  (a)   /Jjt,«(e'!  +  4-<rV       (ft)    /  ja_    a(e*  +  4  -  «-*n 
w\e  +  l       4(e-e-i)    j        WU  +  1        8(e  -  «-•)     ^ 

4.  (a)   («,  *£)  ■    (6)   (,a,  ^)  • 

5.(^,0).  ..(*.«,„).  -^^^Z 

«-(t'°);  9-  (°.  °.  fg)  •  II.  *  =  »=*=l<». 

13.  1TV  in.  from  base  of  cylinder. 

14.  Take  mass  wi  at  origin  and  X-axis  along  the  side  containing  wii  and 

m2  ;  then  x=—  a,  y  =  — — -  a. 
24  24 

Art.  138.     Page  282. 

1.  I  bh*  ;  |  W.  5.  J  MP  ;  1  ft*  8.  '■*#  a3  ;  §J  a2. 

2.  £  bhs  ;  TV  ft2.  6.  |  7ra-3 ;  i  a2.  9.  ^  ^5  .  ^  a2. 

3.  |7ra4;  £a2.  7.  ff»«*;  ff  a2.  10.  ^irfca6;  i«2. 

4.  f  Trfca5 ;  I  a2. 

Art.  139.     Pages  285,  286. 

The  squares  of  the  radi.'  of  gyration  are  : 

6.   (a)  1«2;  (6)  fa2;  (c)  |  +  |' 

8    7       ftg  +  qf-fa 

1      2(a  +  &  -  0 

9'   (a)  k   -3(«  +  6-0 
(6)  Ay>  =  p  _  jjt 

Art.  140.     Pages  287,  288. 

2  Ah 


2. 

/   \    b'1      /rN  ffl2 . 

(a)  -;    (6)  j; 

(c)  i&2;  (d)  ^±-^ 

3 

4. 
5 

i(«i2  +  a22). 

1.  \kb(h2*-hx*).  5.   ^vT^ftfti  7. 

2.  §*ftft*.  6*    |V2^af  8. 


348  ANSWERS 


Miscellaneous  Examples.     Pages  288-290. 

1.    12.022.  6.    fa2.  Q    8  tu* 


2.  fTrafc.  o 

3.  fa2.  7.    7ra^2,  1Q       gxcg 

4.  1 7ra2.  Vm« 


,     a6(27\/3  +  10  7r)  ft    4  7rm5  10        Va262  +  1 

u.   — •  o.   — - — ■  •  ia.   7n 

64  ab  a2 


13.  |aC  +  3(a2+4c2)arctan^..  _«2_Ji«!  +  4;r2/r  ,  a\*\ 

14.  ^(«2-62)*.  23  W-W 

12  (Mi  -  b^s) 


18.  f  (|  +  logMV. 

2  \2  ac/ 


24.    10.814  in. 


16.  £  of  density  at  vertex.  25     c=-( R+  r^  +  rg2>\ • 

17.  f  of  density  at  base.  r  a  4  *     / 

18.  2  7r2a26  ;  4  tt2«6.  29.    «  =  \  OC,  y  =  f  OD,  i  =  £  OA 

Art.  144.     Page  298. 

2.  l_^  +  ^_^+£-  +  ...,     -<x><x<ao. 

2!      4!      0!      8! 

3.  1  +  x  log  a  +  ^lQS  q)2  +  X*(l°*  aY  +  •  •  ■,     -*<«<*, 

2 !  o ! 

4.  l+x  +  ^-^+.-.,     -co<x<oo. 

2        8 

5.  1082+1  +  1  +  ^..-,     -oc<x<co. 

6.  -~f 4-f-f +  f ,     -K.jjl. 

x^     xf_x7      X9 -1'<»<I 

9.    *-•£+  (3»2-2w)^ ,     -oo<0<ao. 

10.  i  +  ^  +  ^  +  2|8  +  ^+...,    _£<,<£. 

o  2  2  2 

ii     1   |  ^  |  S  <H  ,  61 »°     277  S»  "•^•a-S 


ANSWERS  349 

t3  Q   f  5 

13.    a;-|T  +  ^. ,     -oo<x<oo. 

3!        5! 

H.  _(*  +  £i  +  *  +  «J!+.A  o<»<;. 

V2      12      45     2520  /        =        2 

18-    1  +  ^  +  g+^+-'    —'<«<«• 

Art.  145.     Page  301. 

i.  e*(i+A+i2+^+...y 

V  2!     81         ) 

2.  5B"  +  m£'»-  ty  +  ^C^-1)  xm-2^2  +  . . . . 

3.  z6  +  6  xhi  +  15  x4?/2  +  20  xsy*  +  15  x2y*  +  6  xy5  +  y6. 

4.  arcsinx  +        h        +         *»         +  *80 +2*3)  +  ... 

(1-x2)*      2!(l-x2)*      3!(l-x2)* 

5.  log  sin  x  +  h  cot  95 esc2  x  +  —  esc2  x  cot  x . 

2  3 

6.  x3  +  9x2+23x  +  22.  7.    x2  +  5x-ll. 

9.  sin  a  +  (a  -  a)  cos  a  -  ^  ~  a^2  sin  a  -  (x~a^  cos  a  +  ...'! 
2!  3! 

10.    5(x-l)3  +  9(x-l)2+4(x-l)-10; 

5(x  -  3)3  +  39(x  -  3)2  +  100(x  -  3)  +  74. 

H.    (,.i)-(*-l)2  +  Ij^3-(*-l)*  +  ..,   0<x<2. 
SS  o    .  '.  4 

12.    1-*Li^  +  £L^,  (»-«)'  +  ...,   0<a!<2a. 
a        a2  a3  a4 

Art.  147.     Page  305. 

7.  0.52360.  8.    1.31105.  9.   2»Jl/'1_f_*2\. 

10.  f§|  x  +  /T7s  sin  2  x  +  t§?  sin  x  cos8  *  H • 

11.  0.746824. 

Art.  149.  Page  311. 

2.  0.21256.  6.   0.016+. 

3.  11.008.  7.    -  19°  15'  <  x  <  19°  16'. 

4.  3.433987 ;  4.290459.  m2     ma 

5.  0.009804  ;  0.010309.  8-    ElTOr<2T+3T 

9.    (a)  10.472.     (6)  10.50.     Relative  error  =  0.0027. 


350 


ANSWERS 
Art.  150.     Page  313. 


Min.  if  x  =  -  4-  mr. 
4 


2.    No  max.  nor  min. 


4.  Min.  if  x  =  —  log  1  • 

2  a     °p 

5.  Max.  if  x  =  — tanx,  -  <x  <  7r. 
Min.  if  x=  —tan  x,  ~  <x<2  ir. 


6. 

Max.  if  6  : 

=  0; 

min.  if  6  =  tt,  or  arc  cos  Vf. 

7. 

Min.  if  a;  = 

=  0. 

9. 

Max. 

if  x 

=  0. 

8. 

Min.  if  x  = 

si. 

Art.  151. 

10. 
Page  317. 

Max. 

if  x 

=  0. 

1. 

2. 

7.   2. 

13.    0. 

19. 

i 

~5- 

2. 

|. 

8.    -2. 

14.    1. 

20. 

-* 

3. 

J- 

9.    1. 

15.    1. 

21. 

1. 

4. 

log  2. 

6 

10.    log  a.  *• 

16.    1. 

22. 

GO. 

5. 

0. 

11.     00. 

17.   e2. 

23. 

h 

6. 

0. 

12.    0. 

18.    0. 

24. 

h 

Art.  152.     Pages  319,  320. 


Triple  point  at  origin  ;   ^  =  0,  2,  -  2. 
dx 


Conjugate  point  at  origin. 
Triple  point  at  origin. 


4.  Cusp  at  origin. 

5.  Triple  point  at  (0,  2). 


Miscellaneous  Exercises.     Pages  320,  321. 

(a)  Convergent  if  x<I|.  (&)  Convergent  if  -co  <#<oo. 

(c)  Convergent  if  —  go  <£<ao  .        (d)  Convergent  if  x  =£  0. 

5.    1 


6s     2  05      17  fl7 
3       15       315 


2      12 


»t-rl-t 


35  + 


8  a;7 


+ 


6.    2  +  ^+^+^-  +  .... 
4       96      1440 


4. 
9. 

12. 
21. 


l_:^L  +  (3,i2-2n)- . 

2!      V  ;4! 

275  ±  85  ft  +  h2  ±  h* ; 

249.617  ;  258.048  ;  266.511  ;  283.509  ;  292.032  ;  300.563. 

*+!**+ Aa*+H*Mv".  14-  L  15-  *•  16-  2- 


x     x2     x9     3  a;4 
2       3        8" 
af     z«     3x* 

X      2  +  3  +  40  + 


x? 

6    '  24 

1  _  X2  +  £_  _  ^JL  + 
3       45 


INDEX 


(Numbers  refer  to  pages.) 


Acceleration,  61. 

angular,  62,  116. 
Adiabatic  expansion,  work  of,  204. 
Algebraic  functions,  39. 
Angular  acceleration,  62,  116. 

speed,  62,  116. 
Anti-derivatives,  101. 
Approximate  integration,  228. 

value  of  small  errors,  79. 
Approximation  formulas,  308. 

Huygens',  321. 
Approximations  for  circular  arcs,  310. 
Area  under  a  curve,  7. 

representing  definite  integral,  178. 
Areas,  approximate  determination  of, 
224. 

by  double  integration,  257,  259. 

of  plane  curves,  184,  186. 

of  surfaces  of  revolution,  196. 
Astroid,  155. 
Asymptotes,  145. 

curvilinear,  167. 

oblique,  146. 

parallel  to  axes,  147. 
Atmospheric  pressure,  122. 

Beams,  strength  and  stiffness  of,  139. 

Cardioid,  155. 

length  of,  195. 
Catenary,  155. 

length  of,  194. 
Center  of  curvature,  158. 

coordinates  of,  159. 
Center  of  gravity,  273. 
Centroid,  273. 

of  circular  arc,  277. 


Centroid,  of  cylindrical  wedge,  278. 

of  semicircle,  278. 
Centroids,  theorems  concerning,  276. 
Circle  of  curvature,  158. 
Circular  arcs,    approximations    for, 
310. 

functions,  differentiation  of,  83. 
Cissoid,  154. 

Coefficients  of  expansion,  65. 
Compound  interest  law,  122. 
Compressibility,  82. 
Computation    by   means    of    series, 
306. 

of  e  and  ic,  306. 

of  logarithms,  307. 

of  trigonometric  functions,  308. 
Concavity  of  a  curve,  142. 
Cone,  second  moment  of,  284. 
Conjugate  point,  149. 
Conoid,  270. 
Constant,  1. 

of  integration,  102. 
Continuity,  15. 

of   functions  of  several  variables, 

.    231. 
Convergence,  tests  of,  293. 
Convergency  of  series,  291,  293. 
Cubical  parabola,  153. 
Curvature,  157. 

center  of,  158. 

circle  of,  158. 
Curvature,  radius  of,  158. 

of  involutes,  163. 
Curve,  concavity  of,  142. 

length  of,  192,  195. 

slope  of,  47. 

tracing,  152. 


351 


352, 


INDEX 


Curves,  areas  under,  184,  186. 

derived,  94. 

with  given  properties,  113. 
Curvilinear  asymptote,  167. 
Cusp,  149. 
Cycloid,  155. 

evolute  of,  167. 

length  of,  194. 

radius  of  curvature  of,  161. 

Definite  integral  as  limit  of  a  sum, 
173. 

definition  of,  169. 

geometric  representation  of,  178. 

of  discontinuous  function,  179. 

properties  of,  171. 
Deflection  of  beam,  137. 
DeMoivre's  theorem,  321. 
Density,  mean,  271. 
Derivative  as  a  rate,  69. 

conditions  for,  30. 

definition  of,  28. 

interpretation  of,  31. 

of  a  constant,  32. 

of  a  function  of  a  function,  41. 

of  a  product,  34. 

of  a  quotient,  36. 

of  a  sum,  32. 

of  algebraic  functions,  39. 

of  inverse  functions,  42. 

of  /a»,  38. 

total,  232,  237. 
Derivatives,   miscellaneous    applica- 
tions of,  64. 

partial,  232. 

successive,  126. 
Derived  curves,  49. 

function,  29. 
Differential  coefficient,  29,  71. 
Differential  notation,  70. 

partial,  242. 

total,  242. 
Differentials  defined,  70. 

differentiation  with,  75. 

exact,  248. 

inexact,  248. 

kinematic  illustrations  of,  72. 
Differentiation,  defined,  29. 


Differentiation,  logarithmic,  96. 

of  algebraic  functions,  40. 

of  exponential  functions,  94. 

of  implicit  functions,  77,  244. 

of  inverse  trigonometric  functions, 
88-90. 

of  series,  303. 

of  trigonometric  functions,  83-85. 

process  of,  29. 

successive,  126. 

theorems  on,  32. 

with  differentials,  75. 
Discharge  through  orifices,  287. 
Discontinuous  function,  definite  inte- 
gral of,  179. 
Divergency  of  series,  291. 
Double  integral,  255. 

point,  149. 

Efficiency  of  a  screw,  138. 

of  water  turbine,  141. 
Electric  currents,  124. 
Ellipse,  perimeter  of,  304. 

radius  of  curvature  of,  162. 
Ellipsoid,  volume  of,  191. 
Epicycloid,  properties  of,  168. 
Equation  of  normal,  55. 

of  tangent,  55. 
Errors,  approximate  value  of,  79. 
Euler's  theorem,  253. 
Evolutes,  165. 
Expansion  of  functions,  296. 

by  integration  and  differentiation 
of  series,  303. 
Exponential  functions,  92. 
Exponential    functions,    differentia- 
tion of,  94. 

problems  involving,  122. 

Falling  bodies,  6. 
First  moment,  273. 
Flow  of  liquid,  287. 

of  water,  287. 
Formulas,  approximation,  308. 
Function,  definition  of,  2. 

of  several  variables,  231. 

point,  240,  249. 

algebraic,  39. 


INDEX 


353 


Functions,  continuity  of,  15. 

explicit  and  implicit,  39. 

graphs  of,  11. 

inverse,  42. 

monotone,  20. 

multiple- valued,  87. 

notation  of,  10. 

single-valued,  87. 

transcendental,  83. 
Fundamental  integrals,  104. 

Gases,  work  done  by,  203. 
Geometrical  interpretation  of  second 

derivative,  128. 
Geometrical  representation  of  definite 

integral,  178. 
Graphs  of  functions,  11. 
Guldin,  theorems  of,  276. 

Harmonic  alternating  current,  205. 

motion,  118. 
Helix,  length  of,  288. 
Hooke's  law,  202. 
Huygens'  approximation,  321. 
Hyperbola,  radius  of  curvature  of, 

159. 
Hypocycloid  of  four  cusps,  155. 

Implicit  functions,  39. 

differentiation  of,  77,  244. 

successive  differentiation  of,  127. 
Increments,  27. 
Independent  variable,  2. 
Indeterminate  forms,  21,  313. 
Infinite  limits  of  integration,  179. 

series,  291. 
Infinitesimal,  14. 
Infinity,  14. 

Inflexion,  points  of,  144. 
Integral,  defined,  101. 

definite,  169. 
Integrals,  approximate  determination 
of,  223. 

fundamental,  104. 

multiple,  254. 

of  xn  and  tt»,  105. 

table  of,  221. 


Integrand,  101. 

Integration,  approximate,  by  series, 
304.     - 

by  inspection,  105. 

by  parts,  207. 

by  substitution,  108. 

by  trigonometric  substitution,  216. 

general  theorems  on,  102. 

mechanical,  227. 

of  irrational  forms,  213. 

of  rational  fractions,  209. 

of  series,  303. 

of  trigonometric  functions,  218. 

special  methods  of,  207. 

successive,  129. 
Interchange   of  order  of  differentia- 
tion, 236. 
Inverse*  trigonometric  functions,   87. 

differentiation  of,  88-90. 

functions,  42. 

derivative  of,  43. 
Inversion  of  sugar,  123. 
Involutes,  162. 

curvature  of,  163. 
Irrational  integrands,  213. 
Isothermal  expansion,  work  of,  201. 

Kinematic  illustration  of  differentials, 
72. 

La  Rue's  rule  for  radius  of  gyration, 

290. 
Latent  heat  of  expansion,  246. 
Law  of  the  mean,  53. 
Leibnitz'  theorem,  140. 
Lemniscate,  155. 
Length  of  a  curve,  192,  195. 
Limit,  definition  of,  12. 

of  monotone  function,  20. 

of  product,  18. 

of  sum  or  difference,  18. 
Limits,  laws  of,  18. 

of  integration,  169. 
change  of,  172. 
infinite,  179. 
Liquid  pressure,  286. 
Logarithmic  differentiation,  96. 
Logarithms,  computation  of,  307. 


354 


INDEX 


Maclaurin's  expansion,  297. 

series,  remainder  for,  303. 

theorem,  303. 
Mass,  271. 
Maxima  and  minima,  132,  311. 

applications  of,  136. 
Mean  density,  271. 

value,  198. 

value  theorem,  53. 
Mechanical  integration,  227. 
Moment,  first,  273. 

of  inertia,  280. 

second,  280. 
Monotone  functions,  20. 
Motion  in  resisting  medium,  119. 

of  a  projectile,  116. 

rectilinear,  114. 

simple  harmonic,  118. 
Multiple  integrals,  254. 

points,  149. 

Napierian  base  e,  93. 
Newton's  law  of  cooling,  123. 
Node,  149. 
Normal,  equation  of,  55. 

length  of,  57,  60.. 
Notation  for  functions,  10. 

Orifices,  discharge  through,  287. 

Pappus,  theorems  of,  276. 
Parameter,  2. 
Parametric  equations,  47. 
Partial  derivative,  232. 

differential,  242. 
Pelton  water  wheel,  power  of,  139. 
Pendulum,  oscillation  of,  305. 
Physical     interpretation     of    second 

derivative,  128. 
Plane  areas  by  double    integration, 

257,  259. 
Planimeters,  227. 
Point  function,  240,  249. 
Points  of  inflexion,  144. 
Polar  element  of  volume,  265. 

normal,  60. 

subnormal,  60. 


Polar  subtangent,  60. 

tangent,  60. 
Potential  of  a  straight  line,  252. 
Power  series,  295. 
Pressure  of  liquids,  286. 
Probability  curve,  156. 

integral,  305. 
Projectile,  motion  of,  116. 

range  of,  118. 
Proportional  parts,  rule  of,  320. 

Radius  of  curvature,  158. 

parametric  representation,  160. 
Radius  of  gyration,  281. 

of  surface  of  revolution,  290. 

of  system,  284. 
Range  of  projectile,  118,  138. 
Rational  fractions,  integration  of,  209. 
Rectification  of  curves,  192,  195. 
Rectilinear  motion,  114. 
Regnault's  experiments,  86, 
Remainder  for  Maclaurin's  series,  303. 
Rolle's  theorem,  52. 
Roots,  extraction  of,  306. 
Rotation  about  a  fixed  axis,  116. 
Roulettes,  162. 
Routh's  rule  for  radii  of. gyration,  290. 


Secant,  slope  of,  4. 
Second  derivative,  interpretations  of, 
128. 

moment,  280. 
Second  moments  about  parallel  axes, 
283. 

theorems  on,  282. 
Semi-cubical  parabola,  153. 
Series,  convergence  of,  293. 

expansion  in,  296. 

for  log  {x  +  h),  300. 

for  sin  x,  298. 

infinite,  291. 

integration  of,  303. 

Maclaurin's,  297. 

power,  295. 

sum  of,  291. 

Taylor's,  299. 

use  of,  in  computation,  306. 


INDEX 


355 


Simpson's  rules,  224. 
Singular  points,  149,  317. 

analytic  condition  for,  317. 
Slope  of  curve,  47. 

of  secant,  4. 

of  tangent,  31. 
Solid  of  revolution,  volume  of,  186. 
Specific  heat,  65,  246. 

of  superheated  steam,  141. 
Speed,  61. 

angular,  62,  116. 

of  falling  body,  6. 
Springs,  compression  of,  204. 
Standard  integrals,  104. 
Strength  of  beams,  139. 
Subnormal,  57. 
Subtangent,  57. 
Successive  derivatives,  126. 

differentiation    of    implicit    func- 
tions, 127. 

integration,  129. 
Summation,  process,  value  of,  176. 

volumes  determined  by,  190. 
Superheated  steam,  equation  of,  230. 

properties  of,  253. 

specific  heat  of,  141. 
Surface,  area  of,  284. 

of  revolution,  area  of,  196. 

Tacnode,  149. 

Tangent,  equation  of,  55. 

length  of,  57,  60. 

slope  of,  3,  31. 
Tan  ^,  cot  ^,  58. 
Tangential  acceleration,  62. 
Taylor's  expansion,  298. 

theorem,  303. 
Taylor's  series,  remainder  for,  302. 


Tests  for  convergence,  293. 
Theorem  of  mean  value,  53. 
Theorems  on  differentiation,  32. 

of  Pappus  and  Guldin,  276. 

relating  to  centroids,  276. 

relating  to  second  moments,  282. 
Total  derivative,  232,  237. 

differential,  242. 
Transcendental  functions,  83. 
Trigonometric    functions,   computa- 
tion of,  308. 

differentiation  of,  83-85. 

integration  of,  218. 
Triple  integral,  255. 

point,  149. 
Turbine,  efficiency  of,  141. 

Van  der  Waals'  equation,  68,  168. 
Variable,  1. 

dependent,  2. 

independent,  2. 
Velocity  components,  73. 
Volume  of  elliptic  paraboloid,  268. 

of  solid  of  revolution,  188. 
Volumes   by   summation    of    slices, 
190. 

by  triple  integration,  261. 

in  polar  coordinates,  265. 

Wetted  perimeter,  136. 
Witch  of  Agnesi,  154. 

area  of,  181. 
Work  of  expanding  gases,  203. 

or  variable  force,  200. 

represented  by  area,  201. 

Zeuner's  equation    for   superheated 
steam,  230. 


14  DAY  USE 


PJP 


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